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15x^2-32x-40=0
a = 15; b = -32; c = -40;
Δ = b2-4ac
Δ = -322-4·15·(-40)
Δ = 3424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3424}=\sqrt{16*214}=\sqrt{16}*\sqrt{214}=4\sqrt{214}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{214}}{2*15}=\frac{32-4\sqrt{214}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{214}}{2*15}=\frac{32+4\sqrt{214}}{30} $
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