Solve -3=-3x^2+8x

$$- 3 = - 3 x ^ { 2 } + 8 x$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$-3x^{2}+8x=-3$$

Add 3 to both sides.

$$-3x^{2}+8x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=8 ab=-3\times 3=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=9 b=-1$$

Rewrite -3x^{2}+8x+3 as \left(-3x^{2}+9x\right)+\left(-x+3\right).

$$\left(-3x^{2}+9x\right)+\left(-x+3\right)$$

Factor out 3x in -3x^{2}+9x.

$$3x\left(-x+3\right)-x+3$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(3x+1\right)$$

To find equation solutions, solve -x+3=0 and 3x+1=0.

$$x=3 x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

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

$$-3x^{2}+8x=-3$$

Add 3 to both sides.

$$-3x^{2}+8x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-8±\sqrt{64+12\times 3}}{2\left(-3\right)}$$

Multiply 12 times 3.

$$x=\frac{-8±\sqrt{64+36}}{2\left(-3\right)}$$

Add 64 to 36.

$$x=\frac{-8±\sqrt{100}}{2\left(-3\right)}$$

Take the square root of 100.

$$x=\frac{-8±10}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-8±10}{-6}$$

Now solve the equation x=\frac{-8±10}{-6} when ± is plus. Add -8 to 10.

$$x=\frac{2}{-6}$$

Reduce the fraction \frac{2}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{3}$$

Now solve the equation x=\frac{-8±10}{-6} when ± is minus. Subtract 10 from -8.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=-\frac{1}{3} x=3$$

Steps for Completing the Square

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

$$-3x^{2}+8x=-3$$

Divide both sides by -3.

$$\frac{-3x^{2}+8x}{-3}=-\frac{3}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{8}{-3}x=-\frac{3}{-3}$$

Divide 8 by -3.

$$x^{2}-\frac{8}{3}x=-\frac{3}{-3}$$

Divide -3 by -3.

$$x^{2}-\frac{8}{3}x=1$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=1+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=1+\frac{16}{9}$$

Add 1 to \frac{16}{9}.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{25}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{5}{3} x-\frac{4}{3}=-\frac{5}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$x=3 x=-\frac{1}{3}$$