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t^{2}-8t+18=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.
t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 18}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-8\right)±\sqrt{64-4\times 18}}{2}
Square -8.
t=\frac{-\left(-8\right)±\sqrt{64-72}}{2}
Multiply -4 times 18.
t=\frac{-\left(-8\right)±\sqrt{-8}}{2}
Add 64 to -72.
t=\frac{-\left(-8\right)±2\sqrt{2}i}{2}
Take the square root of -8.
t=\frac{8±2\sqrt{2}i}{2}
The opposite of -8 is 8.
t=\frac{8+2\sqrt{2}i}{2}
Now solve the equation t=\frac{8±2\sqrt{2}i}{2} when ± is plus. Add 8 to 2i\sqrt{2}.
t=4+\sqrt{2}i
Divide 8+2i\sqrt{2} by 2.
t=\frac{-2\sqrt{2}i+8}{2}
Now solve the equation t=\frac{8±2\sqrt{2}i}{2} when ± is minus. Subtract 2i\sqrt{2} from 8.
t=-\sqrt{2}i+4
Divide 8-2i\sqrt{2} by 2.
t=4+\sqrt{2}i t=-\sqrt{2}i+4
The equation is now solved.
t^{2}-8t+18=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.
t^{2}-8t+18-18=-18
Subtract 18 from both sides of the equation.
t^{2}-8t=-18
Subtracting 18 from itself leaves 0.
t^{2}-8t+\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.
t^{2}-8t+16=-18+16
Square -4.
t^{2}-8t+16=-2
Add -18 to 16.
\left(t-4\right)^{2}=-2
Factor t^{2}-8t+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(t-4\right)^{2}}=\sqrt{-2}
Take the square root of both sides of the equation.
t-4=\sqrt{2}i t-4=-\sqrt{2}i
Simplify.
t=4+\sqrt{2}i t=-\sqrt{2}i+4
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.
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 = 2
Simplify the expression by subtracting 16 on both sides
u^2 = -2 u = \pm\sqrt{-2} = \pm \sqrt{2}i
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{2}i s = 4 + \sqrt{2}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.