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5(3+x)=x(x-3)-4x
We move all terms to the left:
5(3+x)-(x(x-3)-4x)=0
We add all the numbers together, and all the variables
5(x+3)-(x(x-3)-4x)=0
We multiply parentheses
5x-(x(x-3)-4x)+15=0
We calculate terms in parentheses: -(x(x-3)-4x), so:We get rid of parentheses
x(x-3)-4x
We add all the numbers together, and all the variables
-4x+x(x-3)
We multiply parentheses
x^2-4x-3x
We add all the numbers together, and all the variables
x^2-7x
Back to the equation:
-(x^2-7x)
-x^2+5x+7x+15=0
We add all the numbers together, and all the variables
-1x^2+12x+15=0
a = -1; b = 12; c = +15;
Δ = b2-4ac
Δ = 122-4·(-1)·15
Δ = 204
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{51}}{2*-1}=\frac{-12-2\sqrt{51}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{51}}{2*-1}=\frac{-12+2\sqrt{51}}{-2} $
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