4w^2-48w+63=0

Solution for 4w^2-48w+63=0 equation:

4w^2-48w+63=0

a = 4; b = -48; c = +63;
Δ = b2-4ac
Δ = -482-4·4·63
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-36}{2*4}=\frac{12}{8}
=1+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+36}{2*4}=\frac{84}{8}
=10+1/2
$