x2+4x-15=0, need help solving this equation?

x^2+4x-15=0 is a quadratic equation so you need to factor
(x+_)(x-_)=0 then each factor equals 0 so x must make the brackets equal 0. There may be 2 different values for x.

factors of c that add to b for the second step

hmm let me try something else.

well i dont think that you can factor that, so lets try it this way...

x^2+4x-15=0
x^2+4x=15
x(x+4)=15
x(x+4)/15=1
x²+4x/15=1
x = about 2.36

x=-b+-(sq rt of b^2-4ac)/2a
=-4+-(sq rt of 4^2-4(1)(-15))/2(1)
=-4+-(sqrt of 16+60)/2
=-4+-(sqrt of 76)/2
=-2+-(sqrt of 76)

x=-2 + -(4+15)^2
x=-2-(19)^1/2
x=-2+(19)^1/2

Complete the square

x^2 + 4x = 15 Now take the 4... half that (=2) then double it (=4)
x^2 + 4x +4= 15 +4 Now factor the left side and simplify the right
(x+2)(x+2) = 19 Now rewrite left
(x+2)^2 = 19 Square root both sides
x+2 = sqrt[19] Now subtract 2 from both sides
x = sqrt[19] - 2 or -2 + sqrt[19]

x^2 + 4x - 15 = 0
x^2 + 4x = 15
x^2 + 2x + 2x + 4 = 15 + 4
(x^2 + 2x) + (2x + 4) = 19
x(x + 2) + 2(x + 2) = 19
(x + 2)^2 = 19
x + 2 = ±√19
x = ±√19 - 2

x^2 + 4x - 15 = 0

-4 +/- ((4)^2 - 4(1)(-15))^(1/2) /2

-4 +/- (16 + 60)^(1/2) / 2

-4 +/- (76)^(1/2) / 2

-2 +/ - sqrt(76)/2

well i dont think that you can factor that, so lets try it this way...

x^2+4x-15=0
x^2+4x=15
x(x+4)=15
x=15/(x+4)
x^2=15/4
x=square root of 15/4
so X= the square root of 3

????