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(3x+25)=(2x+10)(2x-25)
We move all terms to the left:
(3x+25)-((2x+10)(2x-25))=0
We get rid of parentheses
3x-((2x+10)(2x-25))+25=0
We multiply parentheses ..
-((+4x^2-50x+20x-250))+3x+25=0
We calculate terms in parentheses: -((+4x^2-50x+20x-250)), so:We add all the numbers together, and all the variables
(+4x^2-50x+20x-250)
We get rid of parentheses
4x^2-50x+20x-250
We add all the numbers together, and all the variables
4x^2-30x-250
Back to the equation:
-(4x^2-30x-250)
3x-(4x^2-30x-250)+25=0
We get rid of parentheses
-4x^2+3x+30x+250+25=0
We add all the numbers together, and all the variables
-4x^2+33x+275=0
a = -4; b = 33; c = +275;
Δ = b2-4ac
Δ = 332-4·(-4)·275
Δ = 5489
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{-(33)-\sqrt{5489}}{2*-4}=\frac{-33-\sqrt{5489}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{5489}}{2*-4}=\frac{-33+\sqrt{5489}}{-8} $
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