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-16x^{2}-8x+150=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(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-16\right)\times 150}}{2\left(-16\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, -8 for b, and 150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-8\right)±\sqrt{64-4\left(-16\right)\times 150}}{2\left(-16\right)}
Square -8.
x=\frac{-\left(-8\right)±\sqrt{64+64\times 150}}{2\left(-16\right)}
Multiply -4 times -16.
x=\frac{-\left(-8\right)±\sqrt{64+9600}}{2\left(-16\right)}
Multiply 64 times 150.
x=\frac{-\left(-8\right)±\sqrt{9664}}{2\left(-16\right)}
Add 64 to 9600.
x=\frac{-\left(-8\right)±8\sqrt{151}}{2\left(-16\right)}
Take the square root of 9664.
x=\frac{8±8\sqrt{151}}{2\left(-16\right)}
The opposite of -8 is 8.
x=\frac{8±8\sqrt{151}}{-32}
Multiply 2 times -16.
x=\frac{8\sqrt{151}+8}{-32}
Now solve the equation x=\frac{8±8\sqrt{151}}{-32} when ± is plus. Add 8 to 8\sqrt{151}.
x=\frac{-\sqrt{151}-1}{4}
Divide 8+8\sqrt{151} by -32.
x=\frac{8-8\sqrt{151}}{-32}
Now solve the equation x=\frac{8±8\sqrt{151}}{-32} when ± is minus. Subtract 8\sqrt{151} from 8.
x=\frac{\sqrt{151}-1}{4}
Divide 8-8\sqrt{151} by -32.
x=\frac{-\sqrt{151}-1}{4} x=\frac{\sqrt{151}-1}{4}
The equation is now solved.
-16x^{2}-8x+150=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.
-16x^{2}-8x+150-150=-150
Subtract 150 from both sides of the equation.
-16x^{2}-8x=-150
Subtracting 150 from itself leaves 0.
\frac{-16x^{2}-8x}{-16}=-\frac{150}{-16}
Divide both sides by -16.
x^{2}+\left(-\frac{8}{-16}\right)x=-\frac{150}{-16}
Dividing by -16 undoes the multiplication by -16.
x^{2}+\frac{1}{2}x=-\frac{150}{-16}
Reduce the fraction \frac{-8}{-16} to lowest terms by extracting and canceling out 8.
x^{2}+\frac{1}{2}x=\frac{75}{8}
Reduce the fraction \frac{-150}{-16} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{75}{8}+\left(\frac{1}{4}\right)^{2}
Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{75}{8}+\frac{1}{16}
Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{151}{16}
Add \frac{75}{8} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{4}\right)^{2}=\frac{151}{16}
Factor x^{2}+\frac{1}{2}x+\frac{1}{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{1}{4}\right)^{2}}=\sqrt{\frac{151}{16}}
Take the square root of both sides of the equation.
x+\frac{1}{4}=\frac{\sqrt{151}}{4} x+\frac{1}{4}=-\frac{\sqrt{151}}{4}
Simplify.
x=\frac{\sqrt{151}-1}{4} x=\frac{-\sqrt{151}-1}{4}
Subtract \frac{1}{4} from both sides of the equation.
x ^ 2 +\frac{1}{2}x -\frac{75}{8} = 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{1}{2} rs = -\frac{75}{8}
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{1}{4} - u s = -\frac{1}{4} + u
Two numbers r and s sum up to -\frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{2} = -\frac{1}{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{1}{4} - u) (-\frac{1}{4} + u) = -\frac{75}{8}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{75}{8}
\frac{1}{16} - u^2 = -\frac{75}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{75}{8}-\frac{1}{16} = -\frac{151}{16}
Simplify the expression by subtracting \frac{1}{16} on both sides
u^2 = \frac{151}{16} u = \pm\sqrt{\frac{151}{16}} = \pm \frac{\sqrt{151}}{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{1}{4} - \frac{\sqrt{151}}{4} = -3.322 s = -\frac{1}{4} + \frac{\sqrt{151}}{4} = 2.822
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.