Solve x^2-8x+10=3x

$${ x }^{ 2 } -8x+10 = 3x$$

Solve for x

$$x=1$$

$$x=10$$

Steps Using Factoring

Subtract 3x from both sides.

$$x^{2}-8x+10-3x=0$$

Combine -8x and -3x to get -11x.

$$x^{2}-11x+10=0$$

To solve the equation, factor x^{2}-11x+10 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-11 ab=10$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-10\right)\left(x-1\right)$$

To find equation solutions, solve x-10=0 and x-1=0.

$$x=10 x=1$$

Steps Using Factoring By Grouping

Subtract 3x from both sides.

$$x^{2}-8x+10-3x=0$$

Combine -8x and -3x to get -11x.

$$x^{2}-11x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 10=10$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite x^{2}-11x+10 as \left(x^{2}-10x\right)+\left(-x+10\right).

$$\left(x^{2}-10x\right)+\left(-x+10\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(x-10\right)-\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-1\right)$$

To find equation solutions, solve x-10=0 and x-1=0.

$$x=10 x=1$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$x^{2}-8x+10-3x=0$$

Combine -8x and -3x to get -11x.

$$x^{2}-11x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 10}}{2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 10}}{2}$$

Multiply -4 times 10.

$$x=\frac{-\left(-11\right)±\sqrt{121-40}}{2}$$

Add 121 to -40.

$$x=\frac{-\left(-11\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-11\right)±9}{2}$$

The opposite of -11 is 11.

$$x=\frac{11±9}{2}$$

Now solve the equation x=\frac{11±9}{2} when ± is plus. Add 11 to 9.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{11±9}{2} when ± is minus. Subtract 9 from 11.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=10 x=1$$

Steps for Completing the Square

Subtract 3x from both sides.

$$x^{2}-8x+10-3x=0$$

Combine -8x and -3x to get -11x.

$$x^{2}-11x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-11x=-10$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-10+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-10+\frac{121}{4}$$

Add -10 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{81}{4}$$

Factor x^{2}-11x+\frac{121}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{9}{2} x-\frac{11}{2}=-\frac{9}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=10 x=1$$