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-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} $
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