3x2-8x-40=x2+6x-4

Solution for 3x2-8x-40=x2+6x-4 equation:

3x^2-8x-40=x2+6x-4

We move all terms to the left:

3x^2-8x-40-(x2+6x-4)=0

We add all the numbers together, and all the variables

3x^2-(+x^2+6x-4)-8x-40=0

We get rid of parentheses

3x^2-x^2-6x-8x+4-40=0

We add all the numbers together, and all the variables

2x^2-14x-36=0

a = 2; b = -14; c = -36;
Δ = b2-4ac
Δ = -142-4·2·(-36)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-22}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+22}{2*2}=\frac{36}{4}
=9
$