Skip to main content
Factor
Tick mark Image
Evaluate
Tick mark Image

Similar Problems from Web Search

Share

-5t^{2}+5t+24=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
t=\frac{-5±\sqrt{5^{2}-4\left(-5\right)\times 24}}{2\left(-5\right)}
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{-5±\sqrt{25-4\left(-5\right)\times 24}}{2\left(-5\right)}
Square 5.
t=\frac{-5±\sqrt{25+20\times 24}}{2\left(-5\right)}
Multiply -4 times -5.
t=\frac{-5±\sqrt{25+480}}{2\left(-5\right)}
Multiply 20 times 24.
t=\frac{-5±\sqrt{505}}{2\left(-5\right)}
Add 25 to 480.
t=\frac{-5±\sqrt{505}}{-10}
Multiply 2 times -5.
t=\frac{\sqrt{505}-5}{-10}
Now solve the equation t=\frac{-5±\sqrt{505}}{-10} when ± is plus. Add -5 to \sqrt{505}.
t=-\frac{\sqrt{505}}{10}+\frac{1}{2}
Divide -5+\sqrt{505} by -10.
t=\frac{-\sqrt{505}-5}{-10}
Now solve the equation t=\frac{-5±\sqrt{505}}{-10} when ± is minus. Subtract \sqrt{505} from -5.
t=\frac{\sqrt{505}}{10}+\frac{1}{2}
Divide -5-\sqrt{505} by -10.
-5t^{2}+5t+24=-5\left(t-\left(-\frac{\sqrt{505}}{10}+\frac{1}{2}\right)\right)\left(t-\left(\frac{\sqrt{505}}{10}+\frac{1}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2}-\frac{\sqrt{505}}{10} for x_{1} and \frac{1}{2}+\frac{\sqrt{505}}{10} for x_{2}.
x ^ 2 -1x -\frac{24}{5} = 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 = 1 rs = -\frac{24}{5}
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 = \frac{1}{2} - u s = \frac{1}{2} + u
Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{2}. 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>
(\frac{1}{2} - u) (\frac{1}{2} + u) = -\frac{24}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{24}{5}
\frac{1}{4} - u^2 = -\frac{24}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{24}{5}-\frac{1}{4} = -\frac{101}{20}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{101}{20} u = \pm\sqrt{\frac{101}{20}} = \pm \frac{\sqrt{101}}{\sqrt{20}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{2} - \frac{\sqrt{101}}{\sqrt{20}} = -1.747 s = \frac{1}{2} + \frac{\sqrt{101}}{\sqrt{20}} = 2.747
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.