3x^2+42x+45=0

Solution for 3x^2+42x+45=0 equation:

3x^2+42x+45=0

a = 3; b = 42; c = +45;
Δ = b2-4ac
Δ = 422-4·3·45
Δ = 1224
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:

$\sqrt{\Delta}=\sqrt{1224}=\sqrt{36*34}=\sqrt{36}*\sqrt{34}=6\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{34}}{2*3}=\frac{-42-6\sqrt{34}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{34}}{2*3}=\frac{-42+6\sqrt{34}}{6}
$