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5x^2+12x-33=0
a = 5; b = 12; c = -33;
Δ = b2-4ac
Δ = 122-4·5·(-33)
Δ = 804
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{201}}{2*5}=\frac{-12-2\sqrt{201}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{201}}{2*5}=\frac{-12+2\sqrt{201}}{10} $
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