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16x^2+16x+3=0
a = 16; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·16·3
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*16}=\frac{-24}{32} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*16}=\frac{-8}{32} =-1/4 $
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