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15x^2-20x-9=0
a = 15; b = -20; c = -9;
Δ = b2-4ac
Δ = -202-4·15·(-9)
Δ = 940
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{940}=\sqrt{4*235}=\sqrt{4}*\sqrt{235}=2\sqrt{235}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{235}}{2*15}=\frac{20-2\sqrt{235}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{235}}{2*15}=\frac{20+2\sqrt{235}}{30} $
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