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8x(2x+1)=0
We multiply parentheses
16x^2+8x=0
a = 16; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·16·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*16}=\frac{-16}{32} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*16}=\frac{0}{32} =0 $
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