How to solve (L)^(2)-9L=-3x+5

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (L)^(2)-9L=-3x+5 | Equatix

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$${ L }^{ 2 } -9L=-3x+5$$

$x=-\frac{L^{2}}{3}+3L+\frac{5}{3}$

Swap sides so that all variable terms are on the left hand side.

$$-3x+5=L^{2}-9L$$

Subtract $5$ from both sides.

$$-3x=L^{2}-9L-5$$

Divide both sides by $-3$.

$$\frac{-3x}{-3}=\frac{L^{2}-9L-5}{-3}$$

Dividing by $-3$ undoes the multiplication by $-3$.

$$x=\frac{L^{2}-9L-5}{-3}$$

Divide $L^{2}-9L-5$ by $-3$.

$$x=-\frac{L^{2}}{3}+3L+\frac{5}{3}$$

$L=\frac{\sqrt{101-12x}+9}{2}$
$L=\frac{-\sqrt{101-12x}+9}{2}$

$L=\frac{\sqrt{101-12x}+9}{2}$
$L=\frac{-\sqrt{101-12x}+9}{2}\text{, }x\leq \frac{101}{12}$