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4x^2-16x=113
We move all terms to the left:
4x^2-16x-(113)=0
a = 4; b = -16; c = -113;
Δ = b2-4ac
Δ = -162-4·4·(-113)
Δ = 2064
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2064}=\sqrt{16*129}=\sqrt{16}*\sqrt{129}=4\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{129}}{2*4}=\frac{16-4\sqrt{129}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{129}}{2*4}=\frac{16+4\sqrt{129}}{8} $
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