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6x+3=8x^2+4x
We move all terms to the left:
6x+3-(8x^2+4x)=0
We get rid of parentheses
-8x^2+6x-4x+3=0
We add all the numbers together, and all the variables
-8x^2+2x+3=0
a = -8; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-8)·3
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-8}=\frac{-12}{-16} =3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-8}=\frac{8}{-16} =-1/2 $
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