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X(4X-5)+2X=5(3X+8)
We move all terms to the left:
X(4X-5)+2X-(5(3X+8))=0
We add all the numbers together, and all the variables
2X+X(4X-5)-(5(3X+8))=0
We multiply parentheses
4X^2+2X-5X-(5(3X+8))=0
We calculate terms in parentheses: -(5(3X+8)), so:We add all the numbers together, and all the variables
5(3X+8)
We multiply parentheses
15X+40
Back to the equation:
-(15X+40)
4X^2-3X-(15X+40)=0
We get rid of parentheses
4X^2-3X-15X-40=0
We add all the numbers together, and all the variables
4X^2-18X-40=0
a = 4; b = -18; c = -40;
Δ = b2-4ac
Δ = -182-4·4·(-40)
Δ = 964
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{241}}{2*4}=\frac{18-2\sqrt{241}}{8} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{241}}{2*4}=\frac{18+2\sqrt{241}}{8} $
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