Solve (-x^2+5x)-(-3x+5)=0

$$(- { x }^{ 2 } +5x)-(-3x+5)=0$$

Solve for x

$$x=\sqrt{11}+4\approx 7.31662479$$

$$x=4-\sqrt{11}\approx 0.68337521$$

Steps Using the Quadratic Formula

To find the opposite of -3x+5, find the opposite of each term.

$$-x^{2}+5x+3x-5=0$$

Combine 5x and 3x to get 8x.

$$-x^{2}+8x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$x=\frac{-8±\sqrt{64-20}}{2\left(-1\right)}$$

Add 64 to -20.

$$x=\frac{-8±\sqrt{44}}{2\left(-1\right)}$$

Take the square root of 44.

$$x=\frac{-8±2\sqrt{11}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±2\sqrt{11}}{-2}$$

Now solve the equation x=\frac{-8±2\sqrt{11}}{-2} when ± is plus. Add -8 to 2\sqrt{11}.

$$x=\frac{2\sqrt{11}-8}{-2}$$

Divide -8+2\sqrt{11} by -2.

$$x=4-\sqrt{11}$$

Now solve the equation x=\frac{-8±2\sqrt{11}}{-2} when ± is minus. Subtract 2\sqrt{11} from -8.

$$x=\frac{-2\sqrt{11}-8}{-2}$$

Divide -8-2\sqrt{11} by -2.

$$x=\sqrt{11}+4$$

The equation is now solved.

$$x=4-\sqrt{11} x=\sqrt{11}+4$$

Steps for Completing the Square

To find the opposite of -3x+5, find the opposite of each term.

$$-x^{2}+5x+3x-5=0$$

Combine 5x and 3x to get 8x.

$$-x^{2}+8x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$-x^{2}+8x=5$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{5}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{5}{-1}$$

Divide 5 by -1.

$$x^{2}-8x=-5$$

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

$$x^{2}-8x+\left(-4\right)^{2}=-5+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-5+16$$

Add -5 to 16.

$$x^{2}-8x+16=11$$

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

$$\left(x-4\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{11}$$

Simplify.

$$x-4=\sqrt{11} x-4=-\sqrt{11}$$

Add 4 to both sides of the equation.

$$x=\sqrt{11}+4 x=4-\sqrt{11}$$