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(9x^2+8x)=(2x^2+3x)
We move all terms to the left:
(9x^2+8x)-((2x^2+3x))=0
We get rid of parentheses
9x^2+8x-((2x^2+3x))=0
We calculate terms in parentheses: -((2x^2+3x)), so:We get rid of parentheses
(2x^2+3x)
We get rid of parentheses
2x^2+3x
Back to the equation:
-(2x^2+3x)
9x^2-2x^2+8x-3x=0
We add all the numbers together, and all the variables
7x^2+5x=0
a = 7; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·7·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*7}=\frac{-10}{14} =-5/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*7}=\frac{0}{14} =0 $
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