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3x^2-9x-15=0
a = 3; b = -9; c = -15;
Δ = b2-4ac
Δ = -92-4·3·(-15)
Δ = 261
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{261}=\sqrt{9*29}=\sqrt{9}*\sqrt{29}=3\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{29}}{2*3}=\frac{9-3\sqrt{29}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{29}}{2*3}=\frac{9+3\sqrt{29}}{6} $
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