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16x^2+16x-41=0
a = 16; b = 16; c = -41;
Δ = b2-4ac
Δ = 162-4·16·(-41)
Δ = 2880
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24\sqrt{5}}{2*16}=\frac{-16-24\sqrt{5}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24\sqrt{5}}{2*16}=\frac{-16+24\sqrt{5}}{32} $
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