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2x-3=2x^2+7x
We move all terms to the left:
2x-3-(2x^2+7x)=0
We get rid of parentheses
-2x^2+2x-7x-3=0
We add all the numbers together, and all the variables
-2x^2-5x-3=0
a = -2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·(-2)·(-3)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*-2}=\frac{4}{-4} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*-2}=\frac{6}{-4} =-1+1/2 $
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