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16x^2+36x+18=0
a = 16; b = 36; c = +18;
Δ = b2-4ac
Δ = 362-4·16·18
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12}{2*16}=\frac{-48}{32} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12}{2*16}=\frac{-24}{32} =-3/4 $
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