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15-16x-4x^2=0
a = -4; b = -16; c = +15;
Δ = b2-4ac
Δ = -162-4·(-4)·15
Δ = 496
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{31}}{2*-4}=\frac{16-4\sqrt{31}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{31}}{2*-4}=\frac{16+4\sqrt{31}}{-8} $
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