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X=-4(X+3)(X+4)
We move all terms to the left:
X-(-4(X+3)(X+4))=0
We multiply parentheses ..
-(-4(+X^2+4X+3X+12))+X=0
We calculate terms in parentheses: -(-4(+X^2+4X+3X+12)), so:We get rid of parentheses
-4(+X^2+4X+3X+12)
We multiply parentheses
-4X^2-16X-12X-48
We add all the numbers together, and all the variables
-4X^2-28X-48
Back to the equation:
-(-4X^2-28X-48)
4X^2+28X+X+48=0
We add all the numbers together, and all the variables
4X^2+29X+48=0
a = 4; b = 29; c = +48;
Δ = b2-4ac
Δ = 292-4·4·48
Δ = 73
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{-(29)-\sqrt{73}}{2*4}=\frac{-29-\sqrt{73}}{8} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{73}}{2*4}=\frac{-29+\sqrt{73}}{8} $
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