4x^2+3x+20x=3

Solution for 4x^2+3x+20x=3 equation:

4x^2+3x+20x=3

We move all terms to the left:

4x^2+3x+20x-(3)=0

We add all the numbers together, and all the variables

4x^2+23x-3=0

a = 4; b = 23; c = -3;
Δ = b2-4ac
Δ = 232-4·4·(-3)
Δ = 577
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{577}}{2*4}=\frac{-23-\sqrt{577}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{577}}{2*4}=\frac{-23+\sqrt{577}}{8}
$