3x^2+10x-54=0

Solution for 3x^2+10x-54=0 equation:

3x^2+10x-54=0

a = 3; b = 10; c = -54;
Δ = b2-4ac
Δ = 102-4·3·(-54)
Δ = 748
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:

$\sqrt{\Delta}=\sqrt{748}=\sqrt{4*187}=\sqrt{4}*\sqrt{187}=2\sqrt{187}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{187}}{2*3}=\frac{-10-2\sqrt{187}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{187}}{2*3}=\frac{-10+2\sqrt{187}}{6}
$