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2x^2-9x-5=-3x+3
We move all terms to the left:
2x^2-9x-5-(-3x+3)=0
We get rid of parentheses
2x^2-9x+3x-3-5=0
We add all the numbers together, and all the variables
2x^2-6x-8=0
a = 2; b = -6; c = -8;
Δ = b2-4ac
Δ = -62-4·2·(-8)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*2}=\frac{-4}{4} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*2}=\frac{16}{4} =4 $
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