4(3x^2-28x+45)=0

Solution for 4(3x^2-28x+45)=0 equation:

4(3x^2-28x+45)=0

We multiply parentheses

12x^2-112x+180=0

a = 12; b = -112; c = +180;
Δ = b2-4ac
Δ = -1122-4·12·180
Δ = 3904
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:

$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-8\sqrt{61}}{2*12}=\frac{112-8\sqrt{61}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+8\sqrt{61}}{2*12}=\frac{112+8\sqrt{61}}{24}
$