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-7x^{2}+5x-4=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(-7\right)\left(-4\right)}}{2\left(-7\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\left(-7\right)\left(-4\right)}}{2\left(-7\right)}
Square 5.
x=\frac{-5±\sqrt{25+28\left(-4\right)}}{2\left(-7\right)}
Multiply -4 times -7.
x=\frac{-5±\sqrt{25-112}}{2\left(-7\right)}
Multiply 28 times -4.
x=\frac{-5±\sqrt{-87}}{2\left(-7\right)}
Add 25 to -112.
x=\frac{-5±\sqrt{87}i}{2\left(-7\right)}
Take the square root of -87.
x=\frac{-5±\sqrt{87}i}{-14}
Multiply 2 times -7.
x=\frac{-5+\sqrt{87}i}{-14}
Now solve the equation x=\frac{-5±\sqrt{87}i}{-14} when ± is plus. Add -5 to i\sqrt{87}.
x=\frac{-\sqrt{87}i+5}{14}
Divide -5+i\sqrt{87} by -14.
x=\frac{-\sqrt{87}i-5}{-14}
Now solve the equation x=\frac{-5±\sqrt{87}i}{-14} when ± is minus. Subtract i\sqrt{87} from -5.
x=\frac{5+\sqrt{87}i}{14}
Divide -5-i\sqrt{87} by -14.
x=\frac{-\sqrt{87}i+5}{14} x=\frac{5+\sqrt{87}i}{14}
The equation is now solved.
-7x^{2}+5x-4=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.
-7x^{2}+5x-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
-7x^{2}+5x=-\left(-4\right)
Subtracting -4 from itself leaves 0.
-7x^{2}+5x=4
Subtract -4 from 0.
\frac{-7x^{2}+5x}{-7}=\frac{4}{-7}
Divide both sides by -7.
x^{2}+\frac{5}{-7}x=\frac{4}{-7}
Dividing by -7 undoes the multiplication by -7.
x^{2}-\frac{5}{7}x=\frac{4}{-7}
Divide 5 by -7.
x^{2}-\frac{5}{7}x=-\frac{4}{7}
Divide 4 by -7.
x^{2}-\frac{5}{7}x+\left(-\frac{5}{14}\right)^{2}=-\frac{4}{7}+\left(-\frac{5}{14}\right)^{2}
Divide -\frac{5}{7}, the coefficient of the x term, by 2 to get -\frac{5}{14}. Then add the square of -\frac{5}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{7}x+\frac{25}{196}=-\frac{4}{7}+\frac{25}{196}
Square -\frac{5}{14} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{7}x+\frac{25}{196}=-\frac{87}{196}
Add -\frac{4}{7} to \frac{25}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{14}\right)^{2}=-\frac{87}{196}
Factor x^{2}-\frac{5}{7}x+\frac{25}{196}. 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}{14}\right)^{2}}=\sqrt{-\frac{87}{196}}
Take the square root of both sides of the equation.
x-\frac{5}{14}=\frac{\sqrt{87}i}{14} x-\frac{5}{14}=-\frac{\sqrt{87}i}{14}
Simplify.
x=\frac{5+\sqrt{87}i}{14} x=\frac{-\sqrt{87}i+5}{14}
Add \frac{5}{14} to both sides of the equation.
x ^ 2 -\frac{5}{7}x +\frac{4}{7} = 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}{7} rs = \frac{4}{7}
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}{14} - u s = \frac{5}{14} + u
Two numbers r and s sum up to \frac{5}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{7} = \frac{5}{14}. 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}{14} - u) (\frac{5}{14} + u) = \frac{4}{7}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{7}
\frac{25}{196} - u^2 = \frac{4}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{4}{7}-\frac{25}{196} = \frac{87}{196}
Simplify the expression by subtracting \frac{25}{196} on both sides
u^2 = -\frac{87}{196} u = \pm\sqrt{-\frac{87}{196}} = \pm \frac{\sqrt{87}}{14}i
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}{14} - \frac{\sqrt{87}}{14}i = 0.357 - 0.666i s = \frac{5}{14} + \frac{\sqrt{87}}{14}i = 0.357 + 0.666i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.