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4a^2+2a-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:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*4}=\frac{-16}{8} =-2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*4}=\frac{12}{8} =1+1/2 $
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