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8x^2=-19x-6
We move all terms to the left:
8x^2-(-19x-6)=0
We get rid of parentheses
8x^2+19x+6=0
a = 8; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·8·6
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*8}=\frac{-32}{16} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*8}=\frac{-6}{16} =-3/8 $
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