Solve for x
x=\frac{\sqrt{26}}{2}-2\approx 0.549509757
x=-\frac{\sqrt{26}}{2}-2\approx -4.549509757
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2x^{2}+8x-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.
x=\frac{-8±\sqrt{8^{2}-4\times 2\left(-5\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8±\sqrt{64-4\times 2\left(-5\right)}}{2\times 2}
Square 8.
x=\frac{-8±\sqrt{64-8\left(-5\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-8±\sqrt{64+40}}{2\times 2}
Multiply -8 times -5.
x=\frac{-8±\sqrt{104}}{2\times 2}
Add 64 to 40.
x=\frac{-8±2\sqrt{26}}{2\times 2}
Take the square root of 104.
x=\frac{-8±2\sqrt{26}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{26}-8}{4}
Now solve the equation x=\frac{-8±2\sqrt{26}}{4} when ± is plus. Add -8 to 2\sqrt{26}.
x=\frac{\sqrt{26}}{2}-2
Divide -8+2\sqrt{26} by 4.
x=\frac{-2\sqrt{26}-8}{4}
Now solve the equation x=\frac{-8±2\sqrt{26}}{4} when ± is minus. Subtract 2\sqrt{26} from -8.
x=-\frac{\sqrt{26}}{2}-2
Divide -8-2\sqrt{26} by 4.
x=\frac{\sqrt{26}}{2}-2 x=-\frac{\sqrt{26}}{2}-2
The equation is now solved.
2x^{2}+8x-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.
2x^{2}+8x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
2x^{2}+8x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
2x^{2}+8x=5
Subtract -5 from 0.
\frac{2x^{2}+8x}{2}=\frac{5}{2}
Divide both sides by 2.
x^{2}+\frac{8}{2}x=\frac{5}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+4x=\frac{5}{2}
Divide 8 by 2.
x^{2}+4x+2^{2}=\frac{5}{2}+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{5}{2}+4
Square 2.
x^{2}+4x+4=\frac{13}{2}
Add \frac{5}{2} to 4.
\left(x+2\right)^{2}=\frac{13}{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{13}{2}}
Take the square root of both sides of the equation.
x+2=\frac{\sqrt{26}}{2} x+2=-\frac{\sqrt{26}}{2}
Simplify.
x=\frac{\sqrt{26}}{2}-2 x=-\frac{\sqrt{26}}{2}-2
Subtract 2 from both sides of the equation.
x ^ 2 +4x -\frac{5}{2} = 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 = -4 rs = -\frac{5}{2}
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 = -2 - u s = -2 + u
Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. 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>
(-2 - u) (-2 + u) = -\frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}
4 - u^2 = -\frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{2}-4 = -\frac{13}{2}
Simplify the expression by subtracting 4 on both sides
u^2 = \frac{13}{2} u = \pm\sqrt{\frac{13}{2}} = \pm \frac{\sqrt{13}}{\sqrt{2}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-2 - \frac{\sqrt{13}}{\sqrt{2}} = -4.550 s = -2 + \frac{\sqrt{13}}{\sqrt{2}} = 0.550
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
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}