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15x^2+246x+180=0
a = 15; b = 246; c = +180;
Δ = b2-4ac
Δ = 2462-4·15·180
Δ = 49716
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{49716}=\sqrt{36*1381}=\sqrt{36}*\sqrt{1381}=6\sqrt{1381}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(246)-6\sqrt{1381}}{2*15}=\frac{-246-6\sqrt{1381}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(246)+6\sqrt{1381}}{2*15}=\frac{-246+6\sqrt{1381}}{30} $
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