20x^2+x-63=0

Solution for 20x^2+x-63=0 equation:

20x^2+x-63=0

a = 20; b = 1; c = -63;
Δ = b2-4ac
Δ = 12-4·20·(-63)
Δ = 5041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5041}=71$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-71}{2*20}=\frac{-72}{40}
=-1+4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+71}{2*20}=\frac{70}{40}
=1+3/4
$