Solve -(x-1)^2+5=2x

$$- ( x - 1 ) ^ { 2 } + 5 = 2 x$$

Solve for x

$$x=2$$

$$x=-2$$

Steps by Finding Square Root

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$-\left(x^{2}-2x+1\right)+5=2x$$

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$-x^{2}+2x-1+5=2x$$

Add -1 and 5 to get 4.

$$-x^{2}+2x+4=2x$$

Subtract 2x from both sides.

$$-x^{2}+2x+4-2x=0$$

Combine 2x and -2x to get 0.

$$-x^{2}+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}=-4$$

Divide both sides by -1.

$$x^{2}=\frac{-4}{-1}$$

Fraction \frac{-4}{-1} can be simplified to 4 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$-\left(x^{2}-2x+1\right)+5=2x$$

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$-x^{2}+2x-1+5=2x$$

Add -1 and 5 to get 4.

$$-x^{2}+2x+4=2x$$

Subtract 2x from both sides.

$$-x^{2}+2x+4-2x=0$$

Combine 2x and -2x to get 0.

$$-x^{2}+4=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{0±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{0±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±4}{-2}$$

Now solve the equation x=\frac{0±4}{-2} when ± is plus. Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{0±4}{-2} when ± is minus. Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$