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5x^2+24x+9=0
a = 5; b = 24; c = +9;
Δ = b2-4ac
Δ = 242-4·5·9
Δ = 396
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{11}}{2*5}=\frac{-24-6\sqrt{11}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{11}}{2*5}=\frac{-24+6\sqrt{11}}{10} $
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