3x^2+18x-26=0

Solution for 3x^2+18x-26=0 equation:

3x^2+18x-26=0

a = 3; b = 18; c = -26;
Δ = b2-4ac
Δ = 182-4·3·(-26)
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{159}}{2*3}=\frac{-18-2\sqrt{159}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{159}}{2*3}=\frac{-18+2\sqrt{159}}{6}
$