2x² -16 x+15=0 can you please help me solve this?

1 or 15/2

question huge style a million : For this equation x^2 + 2*x - 15 = 0 , answer here questions : A. Use ending up the sq. to discover the muse of the equation ! answer huge style a million : The equation x^2 + 2*x - 15 = 0 is already in a*x^2+b*x+c=0 type. because of the fact the linked fee is already arranged in a*x^2+b*x+c=0 type, we get the linked fee of a = a million, b = 2, c = -15. 1A. Use ending up the sq. to discover the muse of the equation ! x^2 + 2*x - 15 = 0 ,divide the two facet with a million So we get x^2 + 2*x - 15 = 0 , The coefficient of x is two we could desire to apply the reality that ( x + q )^2 = x^2 + 2*q*x + q^2 , and anticipate that q = 2/2 = a million by using using that actuality we turn the equation into x^2 + 2*x + a million - sixteen = 0 So we can get ( x + a million )^2 - sixteen = 0 And this is a similar with (( x + a million ) - 4 ) * (( x + a million ) + 4 ) = 0 by using using the associative regulation we get ( x + a million - 4 ) * ( x + a million + 4 ) = 0 Do the addition/subtraction, and we get ( x - 3 ) * ( x + 5 ) = 0 We get following solutions x1 = 3 and x2 = -5

This can not be factarized normalley. So use Quadratic eqation to solve.

2x^2 - 16x + 15 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 2
b = -16
c = 15

x = [16 ±√(256 - 120)]/4
x = [16 ±√136]/4
x = [16 ±√(2^2 * 2 * 17)]/4
x = [16 ±2√(2 * 17)]/4
x = [8 ±√34]/2
x = 4 ±(√34)/2

∴ x = 4 ±(√34)/2

x = [ 16 ± √ (256 - 120 ) ] / 4
x = [ 16 ± √ (136 ) ] / 4
x = [ 16 ± 2√ (34 ) ] / 4
x = 4 ± (1/2)√ 34

2x² -16x + 15 = 0
x² - 8x = -15/2
(x - 4)² = 17/2
x = ±√(17/2) + 4

Answer: x = (±√34 + 8) / 2

this doesn't factorise normally, so you got to use the quadratic equation, which is -b + root of b^2 - 4ac / 2a

That's not a perfect square trinomial. You can't complete the square because of the 2 coefficient in front of the x^2, so you do have to use the quadratic formula, which from standard form ax^2+bx+c=0,
-b+or-[sqrtb^2-4ac]/2a So let's plug in the information:

-(-16)+-[sqrt-16^2-4(2)15]/2(2)
16+-[sqrt256-120]/4
16+-(sqrt136)/4
16+-(11.66)/4
16+-2.915
x= 18.915 or
x=14.915 You just have to plug these values back into your original equation to see if one or both are true.
http://www.purplemath.com/modules/solvquad4.htm

wpf

Just need to use quadratic eqn- x = (-b +- sqrt(b^2-4ac))/2a
I get x = sqrt(34)/2 + 4 & 4 - sqrt(34)/2

This doesn't factor, so you'd have to use the quadratic equation and get the decimals/fractions. I just plugged it into a solver online. Enjoy.

x1 = 6.91547594742265
x2 = 1.0845240525773496