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Steps Using the Quadratic Formula
Steps for Completing the Square
Steps Using the Quadratic Formula
All equations of the form can be solved using the quadratic formula: . The quadratic formula gives two solutions, one when is addition and one when it is subtraction.
This equation is in standard form: . Substitute for , for , and for in the quadratic formula, .
Square .
Multiply times .
Multiply times .
Add to .
Take the square root of .
Multiply times .
Now solve the equation when is plus. Add to .
Divide by .
Now solve the equation when is minus. Subtract from .
Divide by .
The equation is now solved.
Solve for y
Steps for Solving Linear Equation
Subtract from both sides. Anything subtracted from zero gives its negation.
Subtract from both sides.
Subtract from both sides.
Divide both sides by .
Dividing by undoes the multiplication by .
Divide by .
Solve for x (complex solution)

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Similar Problems from Web Search

2x^{2}+8x+8-y=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=\frac{-8±\sqrt{8^{2}-4\times 2\left(8-y\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and -y+8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8±\sqrt{64-4\times 2\left(8-y\right)}}{2\times 2}
Square 8.
x=\frac{-8±\sqrt{64-8\left(8-y\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-8±\sqrt{64+8y-64}}{2\times 2}
Multiply -8 times -y+8.
x=\frac{-8±\sqrt{8y}}{2\times 2}
Add 64 to 8y-64.
x=\frac{-8±2\sqrt{2y}}{2\times 2}
Take the square root of 8y.
x=\frac{-8±2\sqrt{2y}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{2y}-8}{4}
Now solve the equation x=\frac{-8±2\sqrt{2y}}{4} when ± is plus. Add -8 to 2\sqrt{2y}.
x=\frac{\sqrt{2y}}{2}-2
Divide -8+2\sqrt{2y} by 4.
x=\frac{-2\sqrt{2y}-8}{4}
Now solve the equation x=\frac{-8±2\sqrt{2y}}{4} when ± is minus. Subtract 2\sqrt{2y} from -8.
x=-\frac{\sqrt{2y}}{2}-2
Divide -8-2\sqrt{2y} by 4.
x=\frac{\sqrt{2y}}{2}-2 x=-\frac{\sqrt{2y}}{2}-2
The equation is now solved.
2x^{2}+8x+8-y=0
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.
2x^{2}+8x+8-y-\left(8-y\right)=-\left(8-y\right)
Subtract -y+8 from both sides of the equation.
2x^{2}+8x=-\left(8-y\right)
Subtracting -y+8 from itself leaves 0.
2x^{2}+8x=y-8
Subtract -y+8 from 0.
\frac{2x^{2}+8x}{2}=\frac{y-8}{2}
Divide both sides by 2.
x^{2}+\frac{8}{2}x=\frac{y-8}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+4x=\frac{y-8}{2}
Divide 8 by 2.
x^{2}+4x=\frac{y}{2}-4
Divide y-8 by 2.
x^{2}+4x+2^{2}=\frac{y}{2}-4+2^{2}
Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+4x+4=\frac{y}{2}-4+4
Square 2.
x^{2}+4x+4=\frac{y}{2}
Add \frac{y}{2}-4 to 4.
\left(x+2\right)^{2}=\frac{y}{2}
Factor x^{2}+4x+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}.
\sqrt{\left(x+2\right)^{2}}=\sqrt{\frac{y}{2}}
Take the square root of both sides of the equation.
x+2=\frac{\sqrt{2y}}{2} x+2=-\frac{\sqrt{2y}}{2}
Simplify.
x=\frac{\sqrt{2y}}{2}-2 x=-\frac{\sqrt{2y}}{2}-2
Subtract 2 from both sides of the equation.
8x-y+8=-2x^{2}
Subtract 2x^{2} from both sides. Anything subtracted from zero gives its negation.
-y+8=-2x^{2}-8x
Subtract 8x from both sides.
-y=-2x^{2}-8x-8
Subtract 8 from both sides.
\frac{-y}{-1}=-\frac{2\left(x+2\right)^{2}}{-1}
Divide both sides by -1.
y=-\frac{2\left(x+2\right)^{2}}{-1}
Dividing by -1 undoes the multiplication by -1.
y=2\left(x+2\right)^{2}
Divide -2\left(2+x\right)^{2} by -1.
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