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150x^{2}+256x-321=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{-256±\sqrt{256^{2}-4\times 150\left(-321\right)}}{2\times 150}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 150 for a, 256 for b, and -321 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-256±\sqrt{65536-4\times 150\left(-321\right)}}{2\times 150}
Square 256.
x=\frac{-256±\sqrt{65536-600\left(-321\right)}}{2\times 150}
Multiply -4 times 150.
x=\frac{-256±\sqrt{65536+192600}}{2\times 150}
Multiply -600 times -321.
x=\frac{-256±\sqrt{258136}}{2\times 150}
Add 65536 to 192600.
x=\frac{-256±2\sqrt{64534}}{2\times 150}
Take the square root of 258136.
x=\frac{-256±2\sqrt{64534}}{300}
Multiply 2 times 150.
x=\frac{2\sqrt{64534}-256}{300}
Now solve the equation x=\frac{-256±2\sqrt{64534}}{300} when ± is plus. Add -256 to 2\sqrt{64534}.
x=\frac{\sqrt{64534}}{150}-\frac{64}{75}
Divide -256+2\sqrt{64534} by 300.
x=\frac{-2\sqrt{64534}-256}{300}
Now solve the equation x=\frac{-256±2\sqrt{64534}}{300} when ± is minus. Subtract 2\sqrt{64534} from -256.
x=-\frac{\sqrt{64534}}{150}-\frac{64}{75}
Divide -256-2\sqrt{64534} by 300.
x=\frac{\sqrt{64534}}{150}-\frac{64}{75} x=-\frac{\sqrt{64534}}{150}-\frac{64}{75}
The equation is now solved.
150x^{2}+256x-321=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.
150x^{2}+256x-321-\left(-321\right)=-\left(-321\right)
Add 321 to both sides of the equation.
150x^{2}+256x=-\left(-321\right)
Subtracting -321 from itself leaves 0.
150x^{2}+256x=321
Subtract -321 from 0.
\frac{150x^{2}+256x}{150}=\frac{321}{150}
Divide both sides by 150.
x^{2}+\frac{256}{150}x=\frac{321}{150}
Dividing by 150 undoes the multiplication by 150.
x^{2}+\frac{128}{75}x=\frac{321}{150}
Reduce the fraction \frac{256}{150} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{128}{75}x=\frac{107}{50}
Reduce the fraction \frac{321}{150} to lowest terms by extracting and canceling out 3.
x^{2}+\frac{128}{75}x+\left(\frac{64}{75}\right)^{2}=\frac{107}{50}+\left(\frac{64}{75}\right)^{2}
Divide \frac{128}{75}, the coefficient of the x term, by 2 to get \frac{64}{75}. Then add the square of \frac{64}{75} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{128}{75}x+\frac{4096}{5625}=\frac{107}{50}+\frac{4096}{5625}
Square \frac{64}{75} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{128}{75}x+\frac{4096}{5625}=\frac{32267}{11250}
Add \frac{107}{50} to \frac{4096}{5625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{64}{75}\right)^{2}=\frac{32267}{11250}
Factor x^{2}+\frac{128}{75}x+\frac{4096}{5625}. 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{64}{75}\right)^{2}}=\sqrt{\frac{32267}{11250}}
Take the square root of both sides of the equation.
x+\frac{64}{75}=\frac{\sqrt{64534}}{150} x+\frac{64}{75}=-\frac{\sqrt{64534}}{150}
Simplify.
x=\frac{\sqrt{64534}}{150}-\frac{64}{75} x=-\frac{\sqrt{64534}}{150}-\frac{64}{75}
Subtract \frac{64}{75} from both sides of the equation.
x ^ 2 +\frac{128}{75}x -\frac{107}{50} = 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 150
r + s = -\frac{128}{75} rs = -\frac{107}{50}
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{64}{75} - u s = -\frac{64}{75} + u
Two numbers r and s sum up to -\frac{128}{75} exactly when the average of the two numbers is \frac{1}{2}*-\frac{128}{75} = -\frac{64}{75}. 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{64}{75} - u) (-\frac{64}{75} + u) = -\frac{107}{50}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{107}{50}
\frac{4096}{5625} - u^2 = -\frac{107}{50}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{107}{50}-\frac{4096}{5625} = -\frac{32267}{11250}
Simplify the expression by subtracting \frac{4096}{5625} on both sides
u^2 = \frac{32267}{11250} u = \pm\sqrt{\frac{32267}{11250}} = \pm \frac{\sqrt{32267}}{\sqrt{11250}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{64}{75} - \frac{\sqrt{32267}}{\sqrt{11250}} = -2.547 s = -\frac{64}{75} + \frac{\sqrt{32267}}{\sqrt{11250}} = 0.840
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.