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-2x^{2}+4x-7=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{-4±\sqrt{4^{2}-4\left(-2\right)\left(-7\right)}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 4 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\left(-2\right)\left(-7\right)}}{2\left(-2\right)}
Square 4.
x=\frac{-4±\sqrt{16+8\left(-7\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-4±\sqrt{16-56}}{2\left(-2\right)}
Multiply 8 times -7.
x=\frac{-4±\sqrt{-40}}{2\left(-2\right)}
Add 16 to -56.
x=\frac{-4±2\sqrt{10}i}{2\left(-2\right)}
Take the square root of -40.
x=\frac{-4±2\sqrt{10}i}{-4}
Multiply 2 times -2.
x=\frac{-4+2\sqrt{10}i}{-4}
Now solve the equation x=\frac{-4±2\sqrt{10}i}{-4} when ± is plus. Add -4 to 2i\sqrt{10}.
x=-\frac{\sqrt{10}i}{2}+1
Divide -4+2i\sqrt{10} by -4.
x=\frac{-2\sqrt{10}i-4}{-4}
Now solve the equation x=\frac{-4±2\sqrt{10}i}{-4} when ± is minus. Subtract 2i\sqrt{10} from -4.
x=\frac{\sqrt{10}i}{2}+1
Divide -4-2i\sqrt{10} by -4.
x=-\frac{\sqrt{10}i}{2}+1 x=\frac{\sqrt{10}i}{2}+1
The equation is now solved.
-2x^{2}+4x-7=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}+4x-7-\left(-7\right)=-\left(-7\right)
Add 7 to both sides of the equation.
-2x^{2}+4x=-\left(-7\right)
Subtracting -7 from itself leaves 0.
-2x^{2}+4x=7
Subtract -7 from 0.
\frac{-2x^{2}+4x}{-2}=\frac{7}{-2}
Divide both sides by -2.
x^{2}+\frac{4}{-2}x=\frac{7}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}-2x=\frac{7}{-2}
Divide 4 by -2.
x^{2}-2x=-\frac{7}{2}
Divide 7 by -2.
x^{2}-2x+1=-\frac{7}{2}+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=-\frac{5}{2}
Add -\frac{7}{2} to 1.
\left(x-1\right)^{2}=-\frac{5}{2}
Factor x^{2}-2x+1. 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-1\right)^{2}}=\sqrt{-\frac{5}{2}}
Take the square root of both sides of the equation.
x-1=\frac{\sqrt{10}i}{2} x-1=-\frac{\sqrt{10}i}{2}
Simplify.
x=\frac{\sqrt{10}i}{2}+1 x=-\frac{\sqrt{10}i}{2}+1
Add 1 to both sides of the equation.
x ^ 2 -2x +\frac{7}{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 = 2 rs = \frac{7}{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 = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>
(1 - u) (1 + u) = \frac{7}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{2}
1 - u^2 = \frac{7}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{7}{2}-1 = \frac{5}{2}
Simplify the expression by subtracting 1 on both sides
u^2 = -\frac{5}{2} u = \pm\sqrt{-\frac{5}{2}} = \pm \frac{\sqrt{5}}{\sqrt{2}}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \frac{\sqrt{5}}{\sqrt{2}}i = 1 - 1.581i s = 1 + \frac{\sqrt{5}}{\sqrt{2}}i = 1 + 1.581i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.