20w^2+73w+63=0

Solution for 20w^2+73w+63=0 equation:

20w^2+73w+63=0

a = 20; b = 73; c = +63;
Δ = b2-4ac
Δ = 732-4·20·63
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(73)-17}{2*20}=\frac{-90}{40}
=-2+1/4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(73)+17}{2*20}=\frac{-56}{40}
=-1+2/5
$