20x+3=40x^2

Solution for 20x+3=40x^2 equation:

20x+3=40x^2

We move all terms to the left:

20x+3-(40x^2)=0

determiningTheFunctionDomain
-40x^2+20x+3=0

a = -40; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·(-40)·3
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{55}}{2*-40}=\frac{-20-4\sqrt{55}}{-80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{55}}{2*-40}=\frac{-20+4\sqrt{55}}{-80}
$