2x2+20x=10

Solution for 2x2+20x=10 equation:

2x^2+20x=10

We move all terms to the left:

2x^2+20x-(10)=0

a = 2; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·2·(-10)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{30}}{2*2}=\frac{-20-4\sqrt{30}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{30}}{2*2}=\frac{-20+4\sqrt{30}}{4}
$