Solve 2x^2+25=-x^2+10x

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

Solve for x (complex solution)

$$x=\frac{5+5\sqrt{2}i}{3}\approx 1.666666667+2.357022604i$$

$$x=\frac{-5\sqrt{2}i+5}{3}\approx 1.666666667-2.357022604i$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$2x^{2}+25+x^{2}=10x$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+25=10x$$

Subtract 10x from both sides.

$$3x^{2}+25-10x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3x^{2}-10x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 25}}{2\times 3}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 25}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\times 25}}{2\times 3}$$

Multiply -12 times 25.

$$x=\frac{-\left(-10\right)±\sqrt{100-300}}{2\times 3}$$

Add 100 to -300.

$$x=\frac{-\left(-10\right)±\sqrt{-200}}{2\times 3}$$

Take the square root of -200.

$$x=\frac{-\left(-10\right)±10\sqrt{2}i}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±10\sqrt{2}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±10\sqrt{2}i}{6}$$

Now solve the equation x=\frac{10±10\sqrt{2}i}{6} when ± is plus. Add 10 to 10i\sqrt{2}.

$$x=\frac{10+10\sqrt{2}i}{6}$$

Divide 10+10i\sqrt{2} by 6.

$$x=\frac{5+5\sqrt{2}i}{3}$$

Now solve the equation x=\frac{10±10\sqrt{2}i}{6} when ± is minus. Subtract 10i\sqrt{2} from 10.

$$x=\frac{-10\sqrt{2}i+10}{6}$$

Divide 10-10i\sqrt{2} by 6.

$$x=\frac{-5\sqrt{2}i+5}{3}$$

The equation is now solved.

$$x=\frac{5+5\sqrt{2}i}{3} x=\frac{-5\sqrt{2}i+5}{3}$$

Steps for Completing the Square

Add x^{2} to both sides.

$$2x^{2}+25+x^{2}=10x$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+25=10x$$

Subtract 10x from both sides.

$$3x^{2}+25-10x=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-10x=-25$$

Divide both sides by 3.

$$\frac{3x^{2}-10x}{3}=-\frac{25}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{10}{3}x=-\frac{25}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{25}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{25}{3}+\frac{25}{9}$$

Add -\frac{25}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{50}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=-\frac{50}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{-\frac{50}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{5\sqrt{2}i}{3} x-\frac{5}{3}=-\frac{5\sqrt{2}i}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=\frac{5+5\sqrt{2}i}{3} x=\frac{-5\sqrt{2}i+5}{3}$$