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x^{2}+5x-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.
x=\frac{-5±\sqrt{5^{2}-4\left(-2\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\left(-2\right)}}{2}
Square 5.
x=\frac{-5±\sqrt{25+8}}{2}
Multiply -4 times -2.
x=\frac{-5±\sqrt{33}}{2}
Add 25 to 8.
x=\frac{\sqrt{33}-5}{2}
Now solve the equation x=\frac{-5±\sqrt{33}}{2} when ± is plus. Add -5 to \sqrt{33}.
x=\frac{-\sqrt{33}-5}{2}
Now solve the equation x=\frac{-5±\sqrt{33}}{2} when ± is minus. Subtract \sqrt{33} from -5.
x=\frac{\sqrt{33}-5}{2} x=\frac{-\sqrt{33}-5}{2}
The equation is now solved.
x^{2}+5x-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.
x^{2}+5x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
x^{2}+5x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
x^{2}+5x=2
Subtract -2 from 0.
x^{2}+5x+\left(\frac{5}{2}\right)^{2}=2+\left(\frac{5}{2}\right)^{2}
Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+5x+\frac{25}{4}=2+\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+5x+\frac{25}{4}=\frac{33}{4}
Add 2 to \frac{25}{4}.
\left(x+\frac{5}{2}\right)^{2}=\frac{33}{4}
Factor x^{2}+5x+\frac{25}{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+\frac{5}{2}\right)^{2}}=\sqrt{\frac{33}{4}}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\frac{\sqrt{33}}{2} x+\frac{5}{2}=-\frac{\sqrt{33}}{2}
Simplify.
x=\frac{\sqrt{33}-5}{2} x=\frac{-\sqrt{33}-5}{2}
Subtract \frac{5}{2} from both sides of the equation.
x ^ 2 +5x -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.
r + s = -5 rs = -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 = -\frac{5}{2} - u s = -\frac{5}{2} + u
Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{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>
(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{25}{4} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{25}{4} = -\frac{33}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{33}{4} u = \pm\sqrt{\frac{33}{4}} = \pm \frac{\sqrt{33}}{2}
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}{2} - \frac{\sqrt{33}}{2} = -5.372 s = -\frac{5}{2} + \frac{\sqrt{33}}{2} = 0.372
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.