Skip to main content
Solve for y
Tick mark Image
Graph

Similar Problems from Web Search

Share

2y^{2}+5y-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{-5±\sqrt{5^{2}-4\times 2\left(-2\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-5±\sqrt{25-4\times 2\left(-2\right)}}{2\times 2}
Square 5.
y=\frac{-5±\sqrt{25-8\left(-2\right)}}{2\times 2}
Multiply -4 times 2.
y=\frac{-5±\sqrt{25+16}}{2\times 2}
Multiply -8 times -2.
y=\frac{-5±\sqrt{41}}{2\times 2}
Add 25 to 16.
y=\frac{-5±\sqrt{41}}{4}
Multiply 2 times 2.
y=\frac{\sqrt{41}-5}{4}
Now solve the equation y=\frac{-5±\sqrt{41}}{4} when ± is plus. Add -5 to \sqrt{41}.
y=\frac{-\sqrt{41}-5}{4}
Now solve the equation y=\frac{-5±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from -5.
y=\frac{\sqrt{41}-5}{4} y=\frac{-\sqrt{41}-5}{4}
The equation is now solved.
2y^{2}+5y-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.
2y^{2}+5y-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
2y^{2}+5y=-\left(-2\right)
Subtracting -2 from itself leaves 0.
2y^{2}+5y=2
Subtract -2 from 0.
\frac{2y^{2}+5y}{2}=\frac{2}{2}
Divide both sides by 2.
y^{2}+\frac{5}{2}y=\frac{2}{2}
Dividing by 2 undoes the multiplication by 2.
y^{2}+\frac{5}{2}y=1
Divide 2 by 2.
y^{2}+\frac{5}{2}y+\left(\frac{5}{4}\right)^{2}=1+\left(\frac{5}{4}\right)^{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.
y^{2}+\frac{5}{2}y+\frac{25}{16}=1+\frac{25}{16}
Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.
y^{2}+\frac{5}{2}y+\frac{25}{16}=\frac{41}{16}
Add 1 to \frac{25}{16}.
\left(y+\frac{5}{4}\right)^{2}=\frac{41}{16}
Factor y^{2}+\frac{5}{2}y+\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}.
\sqrt{\left(y+\frac{5}{4}\right)^{2}}=\sqrt{\frac{41}{16}}
Take the square root of both sides of the equation.
y+\frac{5}{4}=\frac{\sqrt{41}}{4} y+\frac{5}{4}=-\frac{\sqrt{41}}{4}
Simplify.
y=\frac{\sqrt{41}-5}{4} y=\frac{-\sqrt{41}-5}{4}
Subtract \frac{5}{4} from both sides of the equation.
x ^ 2 +\frac{5}{2}x -1 = 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.This is achieved by dividing both sides of the equation by 2
r + s = -\frac{5}{2} rs = -1
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{5}{4} - u s = -\frac{5}{4} + u
Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{4}. 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{5}{4} - u) (-\frac{5}{4} + u) = -1
To solve for unknown quantity u, substitute these in the product equation rs = -1
\frac{25}{16} - u^2 = -1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -1-\frac{25}{16} = -\frac{41}{16}
Simplify the expression by subtracting \frac{25}{16} on both sides
u^2 = \frac{41}{16} u = \pm\sqrt{\frac{41}{16}} = \pm \frac{\sqrt{41}}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{5}{4} - \frac{\sqrt{41}}{4} = -2.851 s = -\frac{5}{4} + \frac{\sqrt{41}}{4} = 0.351
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.