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-16x^2+10x+290=0
a = -16; b = 10; c = +290;
Δ = b2-4ac
Δ = 102-4·(-16)·290
Δ = 18660
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18660}=\sqrt{4*4665}=\sqrt{4}*\sqrt{4665}=2\sqrt{4665}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{4665}}{2*-16}=\frac{-10-2\sqrt{4665}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{4665}}{2*-16}=\frac{-10+2\sqrt{4665}}{-32} $
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