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2x^{2}-16x-36=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(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\left(-36\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -16 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-16\right)±\sqrt{256-4\times 2\left(-36\right)}}{2\times 2}
Square -16.
x=\frac{-\left(-16\right)±\sqrt{256-8\left(-36\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-16\right)±\sqrt{256+288}}{2\times 2}
Multiply -8 times -36.
x=\frac{-\left(-16\right)±\sqrt{544}}{2\times 2}
Add 256 to 288.
x=\frac{-\left(-16\right)±4\sqrt{34}}{2\times 2}
Take the square root of 544.
x=\frac{16±4\sqrt{34}}{2\times 2}
The opposite of -16 is 16.
x=\frac{16±4\sqrt{34}}{4}
Multiply 2 times 2.
x=\frac{4\sqrt{34}+16}{4}
Now solve the equation x=\frac{16±4\sqrt{34}}{4} when ± is plus. Add 16 to 4\sqrt{34}.
x=\sqrt{34}+4
Divide 16+4\sqrt{34} by 4.
x=\frac{16-4\sqrt{34}}{4}
Now solve the equation x=\frac{16±4\sqrt{34}}{4} when ± is minus. Subtract 4\sqrt{34} from 16.
x=4-\sqrt{34}
Divide 16-4\sqrt{34} by 4.
x=\sqrt{34}+4 x=4-\sqrt{34}
The equation is now solved.
2x^{2}-16x-36=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.
2x^{2}-16x-36-\left(-36\right)=-\left(-36\right)
Add 36 to both sides of the equation.
2x^{2}-16x=-\left(-36\right)
Subtracting -36 from itself leaves 0.
2x^{2}-16x=36
Subtract -36 from 0.
\frac{2x^{2}-16x}{2}=\frac{36}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{16}{2}\right)x=\frac{36}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-8x=\frac{36}{2}
Divide -16 by 2.
x^{2}-8x=18
Divide 36 by 2.
x^{2}-8x+\left(-4\right)^{2}=18+\left(-4\right)^{2}
Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-8x+16=18+16
Square -4.
x^{2}-8x+16=34
Add 18 to 16.
\left(x-4\right)^{2}=34
Factor x^{2}-8x+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-4\right)^{2}}=\sqrt{34}
Take the square root of both sides of the equation.
x-4=\sqrt{34} x-4=-\sqrt{34}
Simplify.
x=\sqrt{34}+4 x=4-\sqrt{34}
Add 4 to both sides of the equation.
x ^ 2 -8x -18 = 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 2
r + s = 8 rs = -18
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 = 4 - u s = 4 + u
Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 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>
(4 - u) (4 + u) = -18
To solve for unknown quantity u, substitute these in the product equation rs = -18
16 - u^2 = -18
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -18-16 = -34
Simplify the expression by subtracting 16 on both sides
u^2 = 34 u = \pm\sqrt{34} = \pm \sqrt{34}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =4 - \sqrt{34} = -1.831 s = 4 + \sqrt{34} = 9.831
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.