3x^2-42x-63=0

Solution for 3x^2-42x-63=0 equation:

3x^2-42x-63=0

a = 3; b = -42; c = -63;
Δ = b2-4ac
Δ = -422-4·3·(-63)
Δ = 2520
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:

$\sqrt{\Delta}=\sqrt{2520}=\sqrt{36*70}=\sqrt{36}*\sqrt{70}=6\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{70}}{2*3}=\frac{42-6\sqrt{70}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{70}}{2*3}=\frac{42+6\sqrt{70}}{6}
$