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8x^2+20x+12=0
a = 8; b = 20; c = +12;
Δ = b2-4ac
Δ = 202-4·8·12
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*8}=\frac{-24}{16} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*8}=\frac{-16}{16} =-1 $
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