Solve for x
x=\sqrt{9541}+75\approx 172.678042568
x=75-\sqrt{9541}\approx -22.678042568
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x^{2}-150x-3916=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(-150\right)±\sqrt{\left(-150\right)^{2}-4\left(-3916\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -150 for b, and -3916 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-150\right)±\sqrt{22500-4\left(-3916\right)}}{2}
Square -150.
x=\frac{-\left(-150\right)±\sqrt{22500+15664}}{2}
Multiply -4 times -3916.
x=\frac{-\left(-150\right)±\sqrt{38164}}{2}
Add 22500 to 15664.
x=\frac{-\left(-150\right)±2\sqrt{9541}}{2}
Take the square root of 38164.
x=\frac{150±2\sqrt{9541}}{2}
The opposite of -150 is 150.
x=\frac{2\sqrt{9541}+150}{2}
Now solve the equation x=\frac{150±2\sqrt{9541}}{2} when ± is plus. Add 150 to 2\sqrt{9541}.
x=\sqrt{9541}+75
Divide 150+2\sqrt{9541} by 2.
x=\frac{150-2\sqrt{9541}}{2}
Now solve the equation x=\frac{150±2\sqrt{9541}}{2} when ± is minus. Subtract 2\sqrt{9541} from 150.
x=75-\sqrt{9541}
Divide 150-2\sqrt{9541} by 2.
x=\sqrt{9541}+75 x=75-\sqrt{9541}
The equation is now solved.
x^{2}-150x-3916=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.
x^{2}-150x-3916-\left(-3916\right)=-\left(-3916\right)
Add 3916 to both sides of the equation.
x^{2}-150x=-\left(-3916\right)
Subtracting -3916 from itself leaves 0.
x^{2}-150x=3916
Subtract -3916 from 0.
x^{2}-150x+\left(-75\right)^{2}=3916+\left(-75\right)^{2}
Divide -150, the coefficient of the x term, by 2 to get -75. Then add the square of -75 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-150x+5625=3916+5625
Square -75.
x^{2}-150x+5625=9541
Add 3916 to 5625.
\left(x-75\right)^{2}=9541
Factor x^{2}-150x+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-75\right)^{2}}=\sqrt{9541}
Take the square root of both sides of the equation.
x-75=\sqrt{9541} x-75=-\sqrt{9541}
Simplify.
x=\sqrt{9541}+75 x=75-\sqrt{9541}
Add 75 to both sides of the equation.
x ^ 2 -150x -3916 = 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 = 150 rs = -3916
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 = 75 - u s = 75 + u
Two numbers r and s sum up to 150 exactly when the average of the two numbers is \frac{1}{2}*150 = 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>
(75 - u) (75 + u) = -3916
To solve for unknown quantity u, substitute these in the product equation rs = -3916
5625 - u^2 = -3916
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -3916-5625 = -9541
Simplify the expression by subtracting 5625 on both sides
u^2 = 9541 u = \pm\sqrt{9541} = \pm \sqrt{9541}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =75 - \sqrt{9541} = -22.678 s = 75 + \sqrt{9541} = 172.678
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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