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16x^2+62x+42=0
a = 16; b = 62; c = +42;
Δ = b2-4ac
Δ = 622-4·16·42
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-34}{2*16}=\frac{-96}{32} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+34}{2*16}=\frac{-28}{32} =-7/8 $
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