Solve for t
t=-3+\sqrt{14}i\approx -3+3.741657387i
t=-\sqrt{14}i-3\approx -3-3.741657387i
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t^{2}+6t+23=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{-6±\sqrt{6^{2}-4\times 23}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-6±\sqrt{36-4\times 23}}{2}
Square 6.
t=\frac{-6±\sqrt{36-92}}{2}
Multiply -4 times 23.
t=\frac{-6±\sqrt{-56}}{2}
Add 36 to -92.
t=\frac{-6±2\sqrt{14}i}{2}
Take the square root of -56.
t=\frac{-6+2\sqrt{14}i}{2}
Now solve the equation t=\frac{-6±2\sqrt{14}i}{2} when ± is plus. Add -6 to 2i\sqrt{14}.
t=-3+\sqrt{14}i
Divide -6+2i\sqrt{14} by 2.
t=\frac{-2\sqrt{14}i-6}{2}
Now solve the equation t=\frac{-6±2\sqrt{14}i}{2} when ± is minus. Subtract 2i\sqrt{14} from -6.
t=-\sqrt{14}i-3
Divide -6-2i\sqrt{14} by 2.
t=-3+\sqrt{14}i t=-\sqrt{14}i-3
The equation is now solved.
t^{2}+6t+23=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}+6t+23-23=-23
Subtract 23 from both sides of the equation.
t^{2}+6t=-23
Subtracting 23 from itself leaves 0.
t^{2}+6t+3^{2}=-23+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}+6t+9=-23+9
Square 3.
t^{2}+6t+9=-14
Add -23 to 9.
\left(t+3\right)^{2}=-14
Factor t^{2}+6t+9. 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+3\right)^{2}}=\sqrt{-14}
Take the square root of both sides of the equation.
t+3=\sqrt{14}i t+3=-\sqrt{14}i
Simplify.
t=-3+\sqrt{14}i t=-\sqrt{14}i-3
Subtract 3 from both sides of the equation.
x ^ 2 +6x +23 = 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 = -6 rs = 23
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 = -3 - u s = -3 + u
Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>
(-3 - u) (-3 + u) = 23
To solve for unknown quantity u, substitute these in the product equation rs = 23
9 - u^2 = 23
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 23-9 = 14
Simplify the expression by subtracting 9 on both sides
u^2 = -14 u = \pm\sqrt{-14} = \pm \sqrt{14}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - \sqrt{14}i s = -3 + \sqrt{14}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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Linear equation
y = 3x + 4
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Matrix
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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