quadric equation of 3x^2+5x-8=0?

3x^2+5x-8=0
(3x+8)(x-1)=0
3x+8=0 => x= -8/3;
x-1=0 => x=1;

Hope it helps
Bye
gilvi

just follow

3x^2+5x-8=0
3x^2+8x-3x-8=0
x(3x+8)-1(3x+8)=0
(3x+8)(x-1)=0
x=1, x=-8/3 answer

good luck

3x^2 + 5x - 8 = 0
3x^2 + 8x - 3x - 8 = 0
(3x^2 + 8x) - (3x + 8) = 0
x(3x + 8) - 1(3x + 8) = 0
(3x + 8)(x - 1) = 0

3x + 8 =0
3x = -8
x = -8/3

x - 1 = 0
x = 1

∴ x = -8/3 , 1

3x^2+5x-8=0
3x^2+8x-3x-8=0
(3x+8)(x-1)=0
3x+8=0 => x= -8/3;
x-1=0 or x=1
its simple split the middle term into two parts.

3x^2+5x-8=0

=> 3x^2 - 3x + 8x - 8 =0
=> 3x(x - 1) + 8(x - 1) = 0
=> (x -1)* (3x + 8) =0

x = 1 or -8/3

3x² + 5x - 8 = 0

ax² + bx + c = 0
r1,r2 = [-b ± √(b²-4ac)]/2a

[-5 ± √(5² - 4(3)(-8))] / 2(3)
(-5 ± √121) / 6
(-5 ± 11 ) / 6

Answer: -8/3, 1

x= {-5 + or - (25+96)^ 1/2} / 6 = -16/6 & +1
CHECK: 3+5-8=0
and 3(-8/3)^2 + 5(-8/3) -8=0,
Generally when you have ax^2 + bx + c =0
then x={ -b + or - (b ^2 - 4ac)^1/2} / 2a,when the equation has real solutions.

3x^2+5x-8=0
3x^2+8x-3x-8=0
x(3x+8)-1(3x+8)=0
(3x+8)(x-1)=0
x=1,-8/3

this factors to (x-1)(3x+8) = 0

so x-1 = 0 and 3x + 8 =0 are solutions

or x = -1 and x =-8/3