Solve for y
y=\frac{\sqrt{11}+1}{5}\approx 0.863324958
y=\frac{1-\sqrt{11}}{5}\approx -0.463324958
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-5y^{2}+2y+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.
y=\frac{-2±\sqrt{2^{2}-4\left(-5\right)\times 2}}{2\left(-5\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 2 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-2±\sqrt{4-4\left(-5\right)\times 2}}{2\left(-5\right)}
Square 2.
y=\frac{-2±\sqrt{4+20\times 2}}{2\left(-5\right)}
Multiply -4 times -5.
y=\frac{-2±\sqrt{4+40}}{2\left(-5\right)}
Multiply 20 times 2.
y=\frac{-2±\sqrt{44}}{2\left(-5\right)}
Add 4 to 40.
y=\frac{-2±2\sqrt{11}}{2\left(-5\right)}
Take the square root of 44.
y=\frac{-2±2\sqrt{11}}{-10}
Multiply 2 times -5.
y=\frac{2\sqrt{11}-2}{-10}
Now solve the equation y=\frac{-2±2\sqrt{11}}{-10} when ± is plus. Add -2 to 2\sqrt{11}.
y=\frac{1-\sqrt{11}}{5}
Divide -2+2\sqrt{11} by -10.
y=\frac{-2\sqrt{11}-2}{-10}
Now solve the equation y=\frac{-2±2\sqrt{11}}{-10} when ± is minus. Subtract 2\sqrt{11} from -2.
y=\frac{\sqrt{11}+1}{5}
Divide -2-2\sqrt{11} by -10.
y=\frac{1-\sqrt{11}}{5} y=\frac{\sqrt{11}+1}{5}
The equation is now solved.
-5y^{2}+2y+2=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.
-5y^{2}+2y+2-2=-2
Subtract 2 from both sides of the equation.
-5y^{2}+2y=-2
Subtracting 2 from itself leaves 0.
\frac{-5y^{2}+2y}{-5}=-\frac{2}{-5}
Divide both sides by -5.
y^{2}+\frac{2}{-5}y=-\frac{2}{-5}
Dividing by -5 undoes the multiplication by -5.
y^{2}-\frac{2}{5}y=-\frac{2}{-5}
Divide 2 by -5.
y^{2}-\frac{2}{5}y=\frac{2}{5}
Divide -2 by -5.
y^{2}-\frac{2}{5}y+\left(-\frac{1}{5}\right)^{2}=\frac{2}{5}+\left(-\frac{1}{5}\right)^{2}
Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-\frac{2}{5}y+\frac{1}{25}=\frac{2}{5}+\frac{1}{25}
Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.
y^{2}-\frac{2}{5}y+\frac{1}{25}=\frac{11}{25}
Add \frac{2}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y-\frac{1}{5}\right)^{2}=\frac{11}{25}
Factor y^{2}-\frac{2}{5}y+\frac{1}{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}.
\sqrt{\left(y-\frac{1}{5}\right)^{2}}=\sqrt{\frac{11}{25}}
Take the square root of both sides of the equation.
y-\frac{1}{5}=\frac{\sqrt{11}}{5} y-\frac{1}{5}=-\frac{\sqrt{11}}{5}
Simplify.
y=\frac{\sqrt{11}+1}{5} y=\frac{1-\sqrt{11}}{5}
Add \frac{1}{5} to both sides of the equation.
x ^ 2 -\frac{2}{5}x -\frac{2}{5} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = \frac{2}{5} rs = -\frac{2}{5}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = \frac{1}{5} - u s = \frac{1}{5} + u
Two numbers r and s sum up to \frac{2}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{5} = \frac{1}{5}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{1}{5} - u) (\frac{1}{5} + u) = -\frac{2}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{5}
\frac{1}{25} - u^2 = -\frac{2}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{5}-\frac{1}{25} = -\frac{11}{25}
Simplify the expression by subtracting \frac{1}{25} on both sides
u^2 = \frac{11}{25} u = \pm\sqrt{\frac{11}{25}} = \pm \frac{\sqrt{11}}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{5} - \frac{\sqrt{11}}{5} = -0.463 s = \frac{1}{5} + \frac{\sqrt{11}}{5} = 0.863
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}