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-16x^2+41x=4
We move all terms to the left:
-16x^2+41x-(4)=0
a = -16; b = 41; c = -4;
Δ = b2-4ac
Δ = 412-4·(-16)·(-4)
Δ = 1425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-5\sqrt{57}}{2*-16}=\frac{-41-5\sqrt{57}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+5\sqrt{57}}{2*-16}=\frac{-41+5\sqrt{57}}{-32} $
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