3x^2+34x+63=0

Solution for 3x^2+34x+63=0 equation:

3x^2+34x+63=0

a = 3; b = 34; c = +63;
Δ = b2-4ac
Δ = 342-4·3·63
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-20}{2*3}=\frac{-54}{6}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+20}{2*3}=\frac{-14}{6}
=-2+1/3
$