20x^2=3x+56

Solution for 20x^2=3x+56 equation:

20x^2=3x+56

We move all terms to the left:

20x^2-(3x+56)=0

We get rid of parentheses

20x^2-3x-56=0

a = 20; b = -3; c = -56;
Δ = b2-4ac
Δ = -32-4·20·(-56)
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-67}{2*20}=\frac{-64}{40}
=-1+3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+67}{2*20}=\frac{70}{40}
=1+3/4
$