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4x^2+12x=154
We move all terms to the left:
4x^2+12x-(154)=0
a = 4; b = 12; c = -154;
Δ = b2-4ac
Δ = 122-4·4·(-154)
Δ = 2608
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2608}=\sqrt{16*163}=\sqrt{16}*\sqrt{163}=4\sqrt{163}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{163}}{2*4}=\frac{-12-4\sqrt{163}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{163}}{2*4}=\frac{-12+4\sqrt{163}}{8} $
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