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12x^2-14x-25=0
a = 12; b = -14; c = -25;
Δ = b2-4ac
Δ = -142-4·12·(-25)
Δ = 1396
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1396}=\sqrt{4*349}=\sqrt{4}*\sqrt{349}=2\sqrt{349}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{349}}{2*12}=\frac{14-2\sqrt{349}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{349}}{2*12}=\frac{14+2\sqrt{349}}{24} $
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