Solve (139)^2+2y^2+20y=494

$$( 139 ) ^ { 2 } + 2 y ^ { 2 } + 20 y = 494$$

Solve for y

$$y=\frac{\sqrt{37554}i}{2}-5\approx -5+96.894272277i$$

$$y=-\frac{\sqrt{37554}i}{2}-5\approx -5-96.894272277i$$

Steps Using the Quadratic Formula

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.

$$2y^{2}+20y+19321=494$$

Subtract 494 from both sides of the equation.

$$2y^{2}+20y+19321-494=494-494$$

Subtracting 494 from itself leaves 0.

$$2y^{2}+20y+19321-494=0$$

Subtract 494 from 19321.

$$2y^{2}+20y+18827=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 20 for b, and 18827 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-20±\sqrt{20^{2}-4\times 2\times 18827}}{2\times 2}$$

Square 20.

$$y=\frac{-20±\sqrt{400-4\times 2\times 18827}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-20±\sqrt{400-8\times 18827}}{2\times 2}$$

Multiply -8 times 18827.

$$y=\frac{-20±\sqrt{400-150616}}{2\times 2}$$

Add 400 to -150616.

$$y=\frac{-20±\sqrt{-150216}}{2\times 2}$$

Take the square root of -150216.

$$y=\frac{-20±2\sqrt{37554}i}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-20±2\sqrt{37554}i}{4}$$

Now solve the equation y=\frac{-20±2\sqrt{37554}i}{4} when ± is plus. Add -20 to 2i\sqrt{37554}.

$$y=\frac{-20+2\sqrt{37554}i}{4}$$

Divide -20+2i\sqrt{37554} by 4.

$$y=\frac{\sqrt{37554}i}{2}-5$$

Now solve the equation y=\frac{-20±2\sqrt{37554}i}{4} when ± is minus. Subtract 2i\sqrt{37554} from -20.

$$y=\frac{-2\sqrt{37554}i-20}{4}$$

Divide -20-2i\sqrt{37554} by 4.

$$y=-\frac{\sqrt{37554}i}{2}-5$$

The equation is now solved.

$$y=\frac{\sqrt{37554}i}{2}-5 y=-\frac{\sqrt{37554}i}{2}-5$$

Steps for Completing the Square

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.

$$2y^{2}+20y+19321=494$$

Subtract 19321 from both sides of the equation.

$$2y^{2}+20y+19321-19321=494-19321$$

Subtracting 19321 from itself leaves 0.

$$2y^{2}+20y=494-19321$$

Subtract 19321 from 494.

$$2y^{2}+20y=-18827$$

Divide both sides by 2.

$$\frac{2y^{2}+20y}{2}=-\frac{18827}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\frac{20}{2}y=-\frac{18827}{2}$$

Divide 20 by 2.

$$y^{2}+10y=-\frac{18827}{2}$$

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.

$$y^{2}+10y+5^{2}=-\frac{18827}{2}+5^{2}$$

Square 5.

$$y^{2}+10y+25=-\frac{18827}{2}+25$$

Add -\frac{18827}{2} to 25.

$$y^{2}+10y+25=-\frac{18777}{2}$$

Factor y^{2}+10y+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}.

$$\left(y+5\right)^{2}=-\frac{18777}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+5\right)^{2}}=\sqrt{-\frac{18777}{2}}$$

Simplify.

$$y+5=\frac{\sqrt{37554}i}{2} y+5=-\frac{\sqrt{37554}i}{2}$$

Subtract 5 from both sides of the equation.

$$y=\frac{\sqrt{37554}i}{2}-5 y=-\frac{\sqrt{37554}i}{2}-5$$