2x^2+4x=8+15

Solution for 2x^2+4x=8+15 equation:

2x^2+4x=8+15

We move all terms to the left:

2x^2+4x-(8+15)=0

We add all the numbers together, and all the variables

2x^2+4x-23=0

a = 2; b = 4; c = -23;
Δ = b2-4ac
Δ = 42-4·2·(-23)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10\sqrt{2}}{2*2}=\frac{-4-10\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10\sqrt{2}}{2*2}=\frac{-4+10\sqrt{2}}{4}
$