2x^2+8x=40-10x

Solution for 2x^2+8x=40-10x equation:

2x^2+8x=40-10x

We move all terms to the left:

2x^2+8x-(40-10x)=0

We add all the numbers together, and all the variables

2x^2+8x-(-10x+40)=0

We get rid of parentheses

2x^2+8x+10x-40=0

We add all the numbers together, and all the variables

2x^2+18x-40=0

a = 2; b = 18; c = -40;
Δ = b2-4ac
Δ = 182-4·2·(-40)
Δ = 644
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{161}}{2*2}=\frac{-18-2\sqrt{161}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{161}}{2*2}=\frac{-18+2\sqrt{161}}{4}
$