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X=(5X+2)(3X-4)
We move all terms to the left:
X-((5X+2)(3X-4))=0
We multiply parentheses ..
-((+15X^2-20X+6X-8))+X=0
We calculate terms in parentheses: -((+15X^2-20X+6X-8)), so:We add all the numbers together, and all the variables
(+15X^2-20X+6X-8)
We get rid of parentheses
15X^2-20X+6X-8
We add all the numbers together, and all the variables
15X^2-14X-8
Back to the equation:
-(15X^2-14X-8)
X-(15X^2-14X-8)=0
We get rid of parentheses
-15X^2+X+14X+8=0
We add all the numbers together, and all the variables
-15X^2+15X+8=0
a = -15; b = 15; c = +8;
Δ = b2-4ac
Δ = 152-4·(-15)·8
Δ = 705
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{-(15)-\sqrt{705}}{2*-15}=\frac{-15-\sqrt{705}}{-30} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{705}}{2*-15}=\frac{-15+\sqrt{705}}{-30} $
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