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-2x^{2}-5x+6=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{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\times 6}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 6}}{2\left(-2\right)}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25+8\times 6}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-5\right)±\sqrt{25+48}}{2\left(-2\right)}
Multiply 8 times 6.
x=\frac{-\left(-5\right)±\sqrt{73}}{2\left(-2\right)}
Add 25 to 48.
x=\frac{5±\sqrt{73}}{2\left(-2\right)}
The opposite of -5 is 5.
x=\frac{5±\sqrt{73}}{-4}
Multiply 2 times -2.
x=\frac{\sqrt{73}+5}{-4}
Now solve the equation x=\frac{5±\sqrt{73}}{-4} when ± is plus. Add 5 to \sqrt{73}.
x=\frac{-\sqrt{73}-5}{4}
Divide 5+\sqrt{73} by -4.
x=\frac{5-\sqrt{73}}{-4}
Now solve the equation x=\frac{5±\sqrt{73}}{-4} when ± is minus. Subtract \sqrt{73} from 5.
x=\frac{\sqrt{73}-5}{4}
Divide 5-\sqrt{73} by -4.
x=\frac{-\sqrt{73}-5}{4} x=\frac{\sqrt{73}-5}{4}
The equation is now solved.
-2x^{2}-5x+6=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}-5x+6-6=-6
Subtract 6 from both sides of the equation.
-2x^{2}-5x=-6
Subtracting 6 from itself leaves 0.
\frac{-2x^{2}-5x}{-2}=-\frac{6}{-2}
Divide both sides by -2.
x^{2}+\left(-\frac{5}{-2}\right)x=-\frac{6}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}+\frac{5}{2}x=-\frac{6}{-2}
Divide -5 by -2.
x^{2}+\frac{5}{2}x=3
Divide -6 by -2.
x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=3+\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.
x^{2}+\frac{5}{2}x+\frac{25}{16}=3+\frac{25}{16}
Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{73}{16}
Add 3 to \frac{25}{16}.
\left(x+\frac{5}{4}\right)^{2}=\frac{73}{16}
Factor x^{2}+\frac{5}{2}x+\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(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{73}{16}}
Take the square root of both sides of the equation.
x+\frac{5}{4}=\frac{\sqrt{73}}{4} x+\frac{5}{4}=-\frac{\sqrt{73}}{4}
Simplify.
x=\frac{\sqrt{73}-5}{4} x=\frac{-\sqrt{73}-5}{4}
Subtract \frac{5}{4} from both sides of the equation.
x ^ 2 +\frac{5}{2}x -3 = 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{5}{2} rs = -3
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) = -3
To solve for unknown quantity u, substitute these in the product equation rs = -3
\frac{25}{16} - u^2 = -3
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -3-\frac{25}{16} = -\frac{73}{16}
Simplify the expression by subtracting \frac{25}{16} on both sides
u^2 = \frac{73}{16} u = \pm\sqrt{\frac{73}{16}} = \pm \frac{\sqrt{73}}{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{73}}{4} = -3.386 s = -\frac{5}{4} + \frac{\sqrt{73}}{4} = 0.886
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.