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6x^2+12x-48=0
a = 6; b = 12; c = -48;
Δ = b2-4ac
Δ = 122-4·6·(-48)
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36}{2*6}=\frac{-48}{12} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36}{2*6}=\frac{24}{12} =2 $
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