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14x^2+16x-17=0
a = 14; b = 16; c = -17;
Δ = b2-4ac
Δ = 162-4·14·(-17)
Δ = 1208
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1208}=\sqrt{4*302}=\sqrt{4}*\sqrt{302}=2\sqrt{302}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{302}}{2*14}=\frac{-16-2\sqrt{302}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{302}}{2*14}=\frac{-16+2\sqrt{302}}{28} $
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