4x+20=5x^2

Solution for 4x+20=5x^2 equation:

4x+20=5x^2

We move all terms to the left:

4x+20-(5x^2)=0

determiningTheFunctionDomain
-5x^2+4x+20=0

a = -5; b = 4; c = +20;
Δ = b2-4ac
Δ = 42-4·(-5)·20
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{26}}{2*-5}=\frac{-4-4\sqrt{26}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{26}}{2*-5}=\frac{-4+4\sqrt{26}}{-10}
$