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-16x^2+20x+290=0
a = -16; b = 20; c = +290;
Δ = b2-4ac
Δ = 202-4·(-16)·290
Δ = 18960
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18960}=\sqrt{16*1185}=\sqrt{16}*\sqrt{1185}=4\sqrt{1185}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{1185}}{2*-16}=\frac{-20-4\sqrt{1185}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{1185}}{2*-16}=\frac{-20+4\sqrt{1185}}{-32} $
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