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(4x-6)(2x+1)=-8x
We move all terms to the left:
(4x-6)(2x+1)-(-8x)=0
We get rid of parentheses
(4x-6)(2x+1)+8x=0
We multiply parentheses ..
(+8x^2+4x-12x-6)+8x=0
We get rid of parentheses
8x^2+4x-12x+8x-6=0
We add all the numbers together, and all the variables
8x^2-6=0
a = 8; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·8·(-6)
Δ = 192
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*8}=\frac{0-8\sqrt{3}}{16} =-\frac{8\sqrt{3}}{16} =-\frac{\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*8}=\frac{0+8\sqrt{3}}{16} =\frac{8\sqrt{3}}{16} =\frac{\sqrt{3}}{2} $
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