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3t^{2}-30t+96=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(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 3\times 96}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -30 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-30\right)±\sqrt{900-4\times 3\times 96}}{2\times 3}
Square -30.
t=\frac{-\left(-30\right)±\sqrt{900-12\times 96}}{2\times 3}
Multiply -4 times 3.
t=\frac{-\left(-30\right)±\sqrt{900-1152}}{2\times 3}
Multiply -12 times 96.
t=\frac{-\left(-30\right)±\sqrt{-252}}{2\times 3}
Add 900 to -1152.
t=\frac{-\left(-30\right)±6\sqrt{7}i}{2\times 3}
Take the square root of -252.
t=\frac{30±6\sqrt{7}i}{2\times 3}
The opposite of -30 is 30.
t=\frac{30±6\sqrt{7}i}{6}
Multiply 2 times 3.
t=\frac{30+6\sqrt{7}i}{6}
Now solve the equation t=\frac{30±6\sqrt{7}i}{6} when ± is plus. Add 30 to 6i\sqrt{7}.
t=5+\sqrt{7}i
Divide 30+6i\sqrt{7} by 6.
t=\frac{-6\sqrt{7}i+30}{6}
Now solve the equation t=\frac{30±6\sqrt{7}i}{6} when ± is minus. Subtract 6i\sqrt{7} from 30.
t=-\sqrt{7}i+5
Divide 30-6i\sqrt{7} by 6.
t=5+\sqrt{7}i t=-\sqrt{7}i+5
The equation is now solved.
3t^{2}-30t+96=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.
3t^{2}-30t+96-96=-96
Subtract 96 from both sides of the equation.
3t^{2}-30t=-96
Subtracting 96 from itself leaves 0.
\frac{3t^{2}-30t}{3}=-\frac{96}{3}
Divide both sides by 3.
t^{2}+\left(-\frac{30}{3}\right)t=-\frac{96}{3}
Dividing by 3 undoes the multiplication by 3.
t^{2}-10t=-\frac{96}{3}
Divide -30 by 3.
t^{2}-10t=-32
Divide -96 by 3.
t^{2}-10t+\left(-5\right)^{2}=-32+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-10t+25=-32+25
Square -5.
t^{2}-10t+25=-7
Add -32 to 25.
\left(t-5\right)^{2}=-7
Factor t^{2}-10t+25. 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-5\right)^{2}}=\sqrt{-7}
Take the square root of both sides of the equation.
t-5=\sqrt{7}i t-5=-\sqrt{7}i
Simplify.
t=5+\sqrt{7}i t=-\sqrt{7}i+5
Add 5 to both sides of the equation.
x ^ 2 -10x +32 = 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 3
r + s = 10 rs = 32
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 = 5 - u s = 5 + u
Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>
(5 - u) (5 + u) = 32
To solve for unknown quantity u, substitute these in the product equation rs = 32
25 - u^2 = 32
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 32-25 = 7
Simplify the expression by subtracting 25 on both sides
u^2 = -7 u = \pm\sqrt{-7} = \pm \sqrt{7}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =5 - \sqrt{7}i s = 5 + \sqrt{7}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.