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16x^2-22x-27=0
a = 16; b = -22; c = -27;
Δ = b2-4ac
Δ = -222-4·16·(-27)
Δ = 2212
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2212}=\sqrt{4*553}=\sqrt{4}*\sqrt{553}=2\sqrt{553}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{553}}{2*16}=\frac{22-2\sqrt{553}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{553}}{2*16}=\frac{22+2\sqrt{553}}{32} $
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