Solve z^2+27=10z

$${ z }^{ 2 } +27 = 10z$$

Solve for z

$$z=5+\sqrt{2}i\approx 5+1.414213562i$$

$$z=-\sqrt{2}i+5\approx 5-1.414213562i$$

Steps Using the Quadratic Formula

Subtract 10z from both sides.

$$z^{2}+27-10z=0$$

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.

$$z^{2}-10z+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 27}}{2}$$

Square -10.

$$z=\frac{-\left(-10\right)±\sqrt{100-4\times 27}}{2}$$

Multiply -4 times 27.

$$z=\frac{-\left(-10\right)±\sqrt{100-108}}{2}$$

Add 100 to -108.

$$z=\frac{-\left(-10\right)±\sqrt{-8}}{2}$$

Take the square root of -8.

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

The opposite of -10 is 10.

$$z=\frac{10±2\sqrt{2}i}{2}$$

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

$$z=\frac{10+2\sqrt{2}i}{2}$$

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

$$z=5+\sqrt{2}i$$

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

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

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

$$z=-\sqrt{2}i+5$$

The equation is now solved.

$$z=5+\sqrt{2}i z=-\sqrt{2}i+5$$

Steps for Completing the Square

Subtract 10z from both sides.

$$z^{2}+27-10z=0$$

Subtract 27 from both sides. Anything subtracted from zero gives its negation.

$$z^{2}-10z=-27$$

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.

$$z^{2}-10z+\left(-5\right)^{2}=-27+\left(-5\right)^{2}$$

Square -5.

$$z^{2}-10z+25=-27+25$$

Add -27 to 25.

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

Factor z^{2}-10z+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$z-5=\sqrt{2}i z-5=-\sqrt{2}i$$

Add 5 to both sides of the equation.

$$z=5+\sqrt{2}i z=-\sqrt{2}i+5$$