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2x^2+6x=21-5x
We move all terms to the left:
2x^2+6x-(21-5x)=0
We add all the numbers together, and all the variables
2x^2+6x-(-5x+21)=0
We get rid of parentheses
2x^2+6x+5x-21=0
We add all the numbers together, and all the variables
2x^2+11x-21=0
a = 2; b = 11; c = -21;
Δ = b2-4ac
Δ = 112-4·2·(-21)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-17}{2*2}=\frac{-28}{4} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+17}{2*2}=\frac{6}{4} =1+1/2 $
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