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16x^2+42x+5=0
a = 16; b = 42; c = +5;
Δ = b2-4ac
Δ = 422-4·16·5
Δ = 1444
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{1444}=38$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-38}{2*16}=\frac{-80}{32} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+38}{2*16}=\frac{-4}{32} =-1/8 $
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