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2x^2-3x-8=3x-4
We move all terms to the left:
2x^2-3x-8-(3x-4)=0
We get rid of parentheses
2x^2-3x-3x+4-8=0
We add all the numbers together, and all the variables
2x^2-6x-4=0
a = 2; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·2·(-4)
Δ = 68
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{17}}{2*2}=\frac{6-2\sqrt{17}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{17}}{2*2}=\frac{6+2\sqrt{17}}{4} $
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