5x^2+40x=10

Solution for 5x^2+40x=10 equation:

5x^2+40x=10

We move all terms to the left:

5x^2+40x-(10)=0

a = 5; b = 40; c = -10;
Δ = b2-4ac
Δ = 402-4·5·(-10)
Δ = 1800
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:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-30\sqrt{2}}{2*5}=\frac{-40-30\sqrt{2}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+30\sqrt{2}}{2*5}=\frac{-40+30\sqrt{2}}{10}
$