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10x^{2}+24x+16=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{-24±\sqrt{24^{2}-4\times 10\times 16}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 24 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-24±\sqrt{576-4\times 10\times 16}}{2\times 10}
Square 24.
x=\frac{-24±\sqrt{576-40\times 16}}{2\times 10}
Multiply -4 times 10.
x=\frac{-24±\sqrt{576-640}}{2\times 10}
Multiply -40 times 16.
x=\frac{-24±\sqrt{-64}}{2\times 10}
Add 576 to -640.
x=\frac{-24±8i}{2\times 10}
Take the square root of -64.
x=\frac{-24±8i}{20}
Multiply 2 times 10.
x=\frac{-24+8i}{20}
Now solve the equation x=\frac{-24±8i}{20} when ± is plus. Add -24 to 8i.
x=-\frac{6}{5}+\frac{2}{5}i
Divide -24+8i by 20.
x=\frac{-24-8i}{20}
Now solve the equation x=\frac{-24±8i}{20} when ± is minus. Subtract 8i from -24.
x=-\frac{6}{5}-\frac{2}{5}i
Divide -24-8i by 20.
x=-\frac{6}{5}+\frac{2}{5}i x=-\frac{6}{5}-\frac{2}{5}i
The equation is now solved.
10x^{2}+24x+16=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.
10x^{2}+24x+16-16=-16
Subtract 16 from both sides of the equation.
10x^{2}+24x=-16
Subtracting 16 from itself leaves 0.
\frac{10x^{2}+24x}{10}=-\frac{16}{10}
Divide both sides by 10.
x^{2}+\frac{24}{10}x=-\frac{16}{10}
Dividing by 10 undoes the multiplication by 10.
x^{2}+\frac{12}{5}x=-\frac{16}{10}
Reduce the fraction \frac{24}{10} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{12}{5}x=-\frac{8}{5}
Reduce the fraction \frac{-16}{10} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=-\frac{8}{5}+\left(\frac{6}{5}\right)^{2}
Divide \frac{12}{5}, the coefficient of the x term, by 2 to get \frac{6}{5}. Then add the square of \frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{12}{5}x+\frac{36}{25}=-\frac{8}{5}+\frac{36}{25}
Square \frac{6}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{12}{5}x+\frac{36}{25}=-\frac{4}{25}
Add -\frac{8}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{6}{5}\right)^{2}=-\frac{4}{25}
Factor x^{2}+\frac{12}{5}x+\frac{36}{25}. 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{6}{5}\right)^{2}}=\sqrt{-\frac{4}{25}}
Take the square root of both sides of the equation.
x+\frac{6}{5}=\frac{2}{5}i x+\frac{6}{5}=-\frac{2}{5}i
Simplify.
x=-\frac{6}{5}+\frac{2}{5}i x=-\frac{6}{5}-\frac{2}{5}i
Subtract \frac{6}{5} from both sides of the equation.
x ^ 2 +\frac{12}{5}x +\frac{8}{5} = 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 10
r + s = -\frac{12}{5} rs = \frac{8}{5}
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{6}{5} - u s = -\frac{6}{5} + u
Two numbers r and s sum up to -\frac{12}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{12}{5} = -\frac{6}{5}. 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{6}{5} - u) (-\frac{6}{5} + u) = \frac{8}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{5}
\frac{36}{25} - u^2 = \frac{8}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{8}{5}-\frac{36}{25} = \frac{4}{25}
Simplify the expression by subtracting \frac{36}{25} on both sides
u^2 = -\frac{4}{25} u = \pm\sqrt{-\frac{4}{25}} = \pm \frac{2}{5}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{6}{5} - \frac{2}{5}i = -1.200 - 0.400i s = -\frac{6}{5} + \frac{2}{5}i = -1.200 + 0.400i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.