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8x^2+4x+6=19
We move all terms to the left:
8x^2+4x+6-(19)=0
We add all the numbers together, and all the variables
8x^2+4x-13=0
a = 8; b = 4; c = -13;
Δ = b2-4ac
Δ = 42-4·8·(-13)
Δ = 432
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12\sqrt{3}}{2*8}=\frac{-4-12\sqrt{3}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12\sqrt{3}}{2*8}=\frac{-4+12\sqrt{3}}{16} $
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