3x^2-8x-31=(2x+3)(x-7)

Solution for 3x^2-8x-31=(2x+3)(x-7) equation:

3x^2-8x-31=(2x+3)(x-7)

We move all terms to the left:

3x^2-8x-31-((2x+3)(x-7))=0

We multiply parentheses ..

3x^2-((+2x^2-14x+3x-21))-8x-31=0


We calculate terms in parentheses: -((+2x^2-14x+3x-21)), so:

(+2x^2-14x+3x-21)

We get rid of parentheses

2x^2-14x+3x-21

We add all the numbers together, and all the variables

2x^2-11x-21

Back to the equation:

-(2x^2-11x-21)


We add all the numbers together, and all the variables

3x^2-8x-(2x^2-11x-21)-31=0

We get rid of parentheses

3x^2-2x^2-8x+11x+21-31=0

We add all the numbers together, and all the variables

x^2+3x-10=0

a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2}
=2
$