4x^2-42x+36=0

Solution for 4x^2-42x+36=0 equation:

4x^2-42x+36=0

a = 4; b = -42; c = +36;
Δ = b2-4ac
Δ = -422-4·4·36
Δ = 1188
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1188}=\sqrt{36*33}=\sqrt{36}*\sqrt{33}=6\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{33}}{2*4}=\frac{42-6\sqrt{33}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{33}}{2*4}=\frac{42+6\sqrt{33}}{8}
$