3x^2+10x-495=0

Solution for 3x^2+10x-495=0 equation:

3x^2+10x-495=0

a = 3; b = 10; c = -495;
Δ = b2-4ac
Δ = 102-4·3·(-495)
Δ = 6040
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:

$\sqrt{\Delta}=\sqrt{6040}=\sqrt{4*1510}=\sqrt{4}*\sqrt{1510}=2\sqrt{1510}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1510}}{2*3}=\frac{-10-2\sqrt{1510}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1510}}{2*3}=\frac{-10+2\sqrt{1510}}{6}
$