Solve b^2-20b+10b=10

$$b ^ { 2 } - 20 b + 10 b = 10$$

Solve for b

$$b=\sqrt{35}+5\approx 10.916079783$$

$$b=5-\sqrt{35}\approx -0.916079783$$

Steps Using the Quadratic Formula

Combine -20b and 10b to get -10b.

$$b^{2}-10b=10$$

Subtract 10 from both sides.

$$b^{2}-10b-10=0$$

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

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

Square -10.

$$b=\frac{-\left(-10\right)±\sqrt{100-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$b=\frac{-\left(-10\right)±\sqrt{100+40}}{2}$$

Add 100 to 40.

$$b=\frac{-\left(-10\right)±\sqrt{140}}{2}$$

Take the square root of 140.

$$b=\frac{-\left(-10\right)±2\sqrt{35}}{2}$$

The opposite of -10 is 10.

$$b=\frac{10±2\sqrt{35}}{2}$$

Now solve the equation b=\frac{10±2\sqrt{35}}{2} when ± is plus. Add 10 to 2\sqrt{35}.

$$b=\frac{2\sqrt{35}+10}{2}$$

Divide 10+2\sqrt{35} by 2.

$$b=\sqrt{35}+5$$

Now solve the equation b=\frac{10±2\sqrt{35}}{2} when ± is minus. Subtract 2\sqrt{35} from 10.

$$b=\frac{10-2\sqrt{35}}{2}$$

Divide 10-2\sqrt{35} by 2.

$$b=5-\sqrt{35}$$

The equation is now solved.

$$b=\sqrt{35}+5 b=5-\sqrt{35}$$

Steps for Completing the Square

Combine -20b and 10b to get -10b.

$$b^{2}-10b=10$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-10b+\left(-5\right)^{2}=10+\left(-5\right)^{2}$$

Square -5.

$$b^{2}-10b+25=10+25$$

Add 10 to 25.

$$b^{2}-10b+25=35$$

Factor b^{2}-10b+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-5\right)^{2}=35$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-5\right)^{2}}=\sqrt{35}$$

Simplify.

$$b-5=\sqrt{35} b-5=-\sqrt{35}$$

Add 5 to both sides of the equation.

$$b=\sqrt{35}+5 b=5-\sqrt{35}$$