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9x^2+18x-53=0
a = 9; b = 18; c = -53;
Δ = b2-4ac
Δ = 182-4·9·(-53)
Δ = 2232
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2232}=\sqrt{36*62}=\sqrt{36}*\sqrt{62}=6\sqrt{62}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{62}}{2*9}=\frac{-18-6\sqrt{62}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{62}}{2*9}=\frac{-18+6\sqrt{62}}{18} $
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