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12x^2+12x-17=0
a = 12; b = 12; c = -17;
Δ = b2-4ac
Δ = 122-4·12·(-17)
Δ = 960
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{15}}{2*12}=\frac{-12-8\sqrt{15}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{15}}{2*12}=\frac{-12+8\sqrt{15}}{24} $
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