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4x+4x^2=78
We move all terms to the left:
4x+4x^2-(78)=0
a = 4; b = 4; c = -78;
Δ = b2-4ac
Δ = 42-4·4·(-78)
Δ = 1264
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1264}=\sqrt{16*79}=\sqrt{16}*\sqrt{79}=4\sqrt{79}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{79}}{2*4}=\frac{-4-4\sqrt{79}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{79}}{2*4}=\frac{-4+4\sqrt{79}}{8} $
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