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(3x*3x-28)=(2x*2x-3x)
We move all terms to the left:
(3x*3x-28)-((2x*2x-3x))=0
We add all the numbers together, and all the variables
(3x*3x-28)-((-3x+2x*2x))=0
We get rid of parentheses
3x*3x-((-3x+2x*2x))-28=0
We calculate terms in parentheses: -((-3x+2x*2x)), so:Wy multiply elements
(-3x+2x*2x)
We get rid of parentheses
-3x+2x*2x
Wy multiply elements
4x^2-3x
Back to the equation:
-(4x^2-3x)
9x^2-(4x^2-3x)-28=0
We get rid of parentheses
9x^2-4x^2+3x-28=0
We add all the numbers together, and all the variables
5x^2+3x-28=0
a = 5; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·5·(-28)
Δ = 569
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{569}}{2*5}=\frac{-3-\sqrt{569}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{569}}{2*5}=\frac{-3+\sqrt{569}}{10} $
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