Skip to main content
Solve for y (complex solution)
Tick mark Image
Solve for y
Tick mark Image
Graph

Similar Problems from Web Search

Share

y^{2}+8y+5=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{-8±\sqrt{8^{2}-4\times 5}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-8±\sqrt{64-4\times 5}}{2}
Square 8.
y=\frac{-8±\sqrt{64-20}}{2}
Multiply -4 times 5.
y=\frac{-8±\sqrt{44}}{2}
Add 64 to -20.
y=\frac{-8±2\sqrt{11}}{2}
Take the square root of 44.
y=\frac{2\sqrt{11}-8}{2}
Now solve the equation y=\frac{-8±2\sqrt{11}}{2} when ± is plus. Add -8 to 2\sqrt{11}.
y=\sqrt{11}-4
Divide -8+2\sqrt{11} by 2.
y=\frac{-2\sqrt{11}-8}{2}
Now solve the equation y=\frac{-8±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from -8.
y=-\sqrt{11}-4
Divide -8-2\sqrt{11} by 2.
y=\sqrt{11}-4 y=-\sqrt{11}-4
The equation is now solved.
y^{2}+8y+5=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.
y^{2}+8y+5-5=-5
Subtract 5 from both sides of the equation.
y^{2}+8y=-5
Subtracting 5 from itself leaves 0.
y^{2}+8y+4^{2}=-5+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+8y+16=-5+16
Square 4.
y^{2}+8y+16=11
Add -5 to 16.
\left(y+4\right)^{2}=11
Factor y^{2}+8y+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+4\right)^{2}}=\sqrt{11}
Take the square root of both sides of the equation.
y+4=\sqrt{11} y+4=-\sqrt{11}
Simplify.
y=\sqrt{11}-4 y=-\sqrt{11}-4
Subtract 4 from both sides of the equation.
x ^ 2 +8x +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 = -8 rs = 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 = -4 - u s = -4 + u
Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -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>
(-4 - u) (-4 + u) = 5
To solve for unknown quantity u, substitute these in the product equation rs = 5
16 - u^2 = 5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 5-16 = -11
Simplify the expression by subtracting 16 on both sides
u^2 = 11 u = \pm\sqrt{11} = \pm \sqrt{11}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-4 - \sqrt{11} = -7.317 s = -4 + \sqrt{11} = -0.683
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
y^{2}+8y+5=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{-8±\sqrt{8^{2}-4\times 5}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-8±\sqrt{64-4\times 5}}{2}
Square 8.
y=\frac{-8±\sqrt{64-20}}{2}
Multiply -4 times 5.
y=\frac{-8±\sqrt{44}}{2}
Add 64 to -20.
y=\frac{-8±2\sqrt{11}}{2}
Take the square root of 44.
y=\frac{2\sqrt{11}-8}{2}
Now solve the equation y=\frac{-8±2\sqrt{11}}{2} when ± is plus. Add -8 to 2\sqrt{11}.
y=\sqrt{11}-4
Divide -8+2\sqrt{11} by 2.
y=\frac{-2\sqrt{11}-8}{2}
Now solve the equation y=\frac{-8±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from -8.
y=-\sqrt{11}-4
Divide -8-2\sqrt{11} by 2.
y=\sqrt{11}-4 y=-\sqrt{11}-4
The equation is now solved.
y^{2}+8y+5=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.
y^{2}+8y+5-5=-5
Subtract 5 from both sides of the equation.
y^{2}+8y=-5
Subtracting 5 from itself leaves 0.
y^{2}+8y+4^{2}=-5+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+8y+16=-5+16
Square 4.
y^{2}+8y+16=11
Add -5 to 16.
\left(y+4\right)^{2}=11
Factor y^{2}+8y+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+4\right)^{2}}=\sqrt{11}
Take the square root of both sides of the equation.
y+4=\sqrt{11} y+4=-\sqrt{11}
Simplify.
y=\sqrt{11}-4 y=-\sqrt{11}-4
Subtract 4 from both sides of the equation.