Solve x^2+65^2=150-65-x

$${ x }^{ 2 } + { 65 }^{ 2 } =150-65-x$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{16559}i}{2}\approx -0.5+64.340889021i$$

$$x=\frac{-\sqrt{16559}i-1}{2}\approx -0.5-64.340889021i$$

Steps Using the Quadratic Formula

Calculate 65 to the power of 2 and get 4225.

$$x^{2}+4225=150-65-x$$

Subtract 65 from 150 to get 85.

$$x^{2}+4225=85-x$$

Subtract 85 from both sides.

$$x^{2}+4225-85=-x$$

Subtract 85 from 4225 to get 4140.

$$x^{2}+4140=-x$$

Add x to both sides.

$$x^{2}+4140+x=0$$

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.

$$x^{2}+x+4140=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and 4140 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4140}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 4140}}{2}$$

Multiply -4 times 4140.

$$x=\frac{-1±\sqrt{1-16560}}{2}$$

Add 1 to -16560.

$$x=\frac{-1±\sqrt{-16559}}{2}$$

Take the square root of -16559.

$$x=\frac{-1±\sqrt{16559}i}{2}$$

Now solve the equation x=\frac{-1±\sqrt{16559}i}{2} when ± is plus. Add -1 to i\sqrt{16559}.

$$x=\frac{-1+\sqrt{16559}i}{2}$$

Now solve the equation x=\frac{-1±\sqrt{16559}i}{2} when ± is minus. Subtract i\sqrt{16559} from -1.

$$x=\frac{-\sqrt{16559}i-1}{2}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{16559}i}{2} x=\frac{-\sqrt{16559}i-1}{2}$$

Steps for Completing the Square

Calculate 65 to the power of 2 and get 4225.

$$x^{2}+4225=150-65-x$$

Subtract 65 from 150 to get 85.

$$x^{2}+4225=85-x$$

Add x to both sides.

$$x^{2}+4225+x=85$$

Subtract 4225 from both sides.

$$x^{2}+x=85-4225$$

Subtract 4225 from 85 to get -4140.

$$x^{2}+x=-4140$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-4140+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=-4140+\frac{1}{4}$$

Add -4140 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=-\frac{16559}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{16559}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{16559}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{16559}i}{2} x+\frac{1}{2}=-\frac{\sqrt{16559}i}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{-1+\sqrt{16559}i}{2} x=\frac{-\sqrt{16559}i-1}{2}$$