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4x^2=-12x+4
We move all terms to the left:
4x^2-(-12x+4)=0
We get rid of parentheses
4x^2+12x-4=0
a = 4; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·4·(-4)
Δ = 208
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{13}}{2*4}=\frac{-12-4\sqrt{13}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{13}}{2*4}=\frac{-12+4\sqrt{13}}{8} $
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