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6x^2+32x+42=0
a = 6; b = 32; c = +42;
Δ = b2-4ac
Δ = 322-4·6·42
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4}{2*6}=\frac{-36}{12} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4}{2*6}=\frac{-28}{12} =-2+1/3 $
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