Solve (u-2)=2u^2+6u+25

$$( u - 2 ) = 2 u ^ { 2 } + 6 u + 25$$

Solve for u

$$u=\frac{-\sqrt{191}i-5}{4}\approx -1.25-3.45506874i$$

$$u=\frac{-5+\sqrt{191}i}{4}\approx -1.25+3.45506874i$$

Steps Using the Quadratic Formula

Subtract 2u^{2} from both sides.

$$u-2-2u^{2}=6u+25$$

Subtract 6u from both sides.

$$u-2-2u^{2}-6u=25$$

Combine u and -6u to get -5u.

$$-5u-2-2u^{2}=25$$

Subtract 25 from both sides.

$$-5u-2-2u^{2}-25=0$$

Subtract 25 from -2 to get -27.

$$-5u-27-2u^{2}=0$$

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.

$$-2u^{2}-5u-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\left(-27\right)}}{2\left(-2\right)}$$

Square -5.

$$u=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\left(-27\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$u=\frac{-\left(-5\right)±\sqrt{25+8\left(-27\right)}}{2\left(-2\right)}$$

Multiply 8 times -27.

$$u=\frac{-\left(-5\right)±\sqrt{25-216}}{2\left(-2\right)}$$

Add 25 to -216.

$$u=\frac{-\left(-5\right)±\sqrt{-191}}{2\left(-2\right)}$$

Take the square root of -191.

$$u=\frac{-\left(-5\right)±\sqrt{191}i}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$u=\frac{5±\sqrt{191}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$u=\frac{5±\sqrt{191}i}{-4}$$

Now solve the equation u=\frac{5±\sqrt{191}i}{-4} when ± is plus. Add 5 to i\sqrt{191}.

$$u=\frac{5+\sqrt{191}i}{-4}$$

Divide 5+i\sqrt{191} by -4.

$$u=\frac{-\sqrt{191}i-5}{4}$$

Now solve the equation u=\frac{5±\sqrt{191}i}{-4} when ± is minus. Subtract i\sqrt{191} from 5.

$$u=\frac{-\sqrt{191}i+5}{-4}$$

Divide 5-i\sqrt{191} by -4.

$$u=\frac{-5+\sqrt{191}i}{4}$$

The equation is now solved.

$$u=\frac{-\sqrt{191}i-5}{4} u=\frac{-5+\sqrt{191}i}{4}$$

Steps for Completing the Square

Subtract 2u^{2} from both sides.

$$u-2-2u^{2}=6u+25$$

Subtract 6u from both sides.

$$u-2-2u^{2}-6u=25$$

Combine u and -6u to get -5u.

$$-5u-2-2u^{2}=25$$

Add 2 to both sides.

$$-5u-2u^{2}=25+2$$

Add 25 and 2 to get 27.

$$-5u-2u^{2}=27$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2u^{2}-5u=27$$

Divide both sides by -2.

$$\frac{-2u^{2}-5u}{-2}=\frac{27}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$u^{2}+\left(-\frac{5}{-2}\right)u=\frac{27}{-2}$$

Divide -5 by -2.

$$u^{2}+\frac{5}{2}u=\frac{27}{-2}$$

Divide 27 by -2.

$$u^{2}+\frac{5}{2}u=-\frac{27}{2}$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$u^{2}+\frac{5}{2}u+\left(\frac{5}{4}\right)^{2}=-\frac{27}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$u^{2}+\frac{5}{2}u+\frac{25}{16}=-\frac{27}{2}+\frac{25}{16}$$

Add -\frac{27}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$u^{2}+\frac{5}{2}u+\frac{25}{16}=-\frac{191}{16}$$

Factor u^{2}+\frac{5}{2}u+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(u+\frac{5}{4}\right)^{2}=-\frac{191}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u+\frac{5}{4}\right)^{2}}=\sqrt{-\frac{191}{16}}$$

Simplify.

$$u+\frac{5}{4}=\frac{\sqrt{191}i}{4} u+\frac{5}{4}=-\frac{\sqrt{191}i}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$u=\frac{-5+\sqrt{191}i}{4} u=\frac{-\sqrt{191}i-5}{4}$$