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4x^2+2x=12
We move all terms to the left:
4x^2+2x-(12)=0
a = 4; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·4·(-12)
Δ = 196
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*4}=\frac{-16}{8} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*4}=\frac{12}{8} =1+1/2 $
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