4x+20x^2=10

Solution for 4x+20x^2=10 equation:

4x+20x^2=10

We move all terms to the left:

4x+20x^2-(10)=0

a = 20; b = 4; c = -10;
Δ = b2-4ac
Δ = 42-4·20·(-10)
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{51}}{2*20}=\frac{-4-4\sqrt{51}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{51}}{2*20}=\frac{-4+4\sqrt{51}}{40}
$