Solve for t
t = \frac{\sqrt{41} + 21}{10} \approx 2.740312424
t = \frac{21 - \sqrt{41}}{10} \approx 1.459687576
Share
Copied to clipboard
5t^{2}-21t+20=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(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 5\times 20}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -21 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-21\right)±\sqrt{441-4\times 5\times 20}}{2\times 5}
Square -21.
t=\frac{-\left(-21\right)±\sqrt{441-20\times 20}}{2\times 5}
Multiply -4 times 5.
t=\frac{-\left(-21\right)±\sqrt{441-400}}{2\times 5}
Multiply -20 times 20.
t=\frac{-\left(-21\right)±\sqrt{41}}{2\times 5}
Add 441 to -400.
t=\frac{21±\sqrt{41}}{2\times 5}
The opposite of -21 is 21.
t=\frac{21±\sqrt{41}}{10}
Multiply 2 times 5.
t=\frac{\sqrt{41}+21}{10}
Now solve the equation t=\frac{21±\sqrt{41}}{10} when ± is plus. Add 21 to \sqrt{41}.
t=\frac{21-\sqrt{41}}{10}
Now solve the equation t=\frac{21±\sqrt{41}}{10} when ± is minus. Subtract \sqrt{41} from 21.
t=\frac{\sqrt{41}+21}{10} t=\frac{21-\sqrt{41}}{10}
The equation is now solved.
5t^{2}-21t+20=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.
5t^{2}-21t+20-20=-20
Subtract 20 from both sides of the equation.
5t^{2}-21t=-20
Subtracting 20 from itself leaves 0.
\frac{5t^{2}-21t}{5}=-\frac{20}{5}
Divide both sides by 5.
t^{2}-\frac{21}{5}t=-\frac{20}{5}
Dividing by 5 undoes the multiplication by 5.
t^{2}-\frac{21}{5}t=-4
Divide -20 by 5.
t^{2}-\frac{21}{5}t+\left(-\frac{21}{10}\right)^{2}=-4+\left(-\frac{21}{10}\right)^{2}
Divide -\frac{21}{5}, the coefficient of the x term, by 2 to get -\frac{21}{10}. Then add the square of -\frac{21}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{21}{5}t+\frac{441}{100}=-4+\frac{441}{100}
Square -\frac{21}{10} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{21}{5}t+\frac{441}{100}=\frac{41}{100}
Add -4 to \frac{441}{100}.
\left(t-\frac{21}{10}\right)^{2}=\frac{41}{100}
Factor t^{2}-\frac{21}{5}t+\frac{441}{100}. 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-\frac{21}{10}\right)^{2}}=\sqrt{\frac{41}{100}}
Take the square root of both sides of the equation.
t-\frac{21}{10}=\frac{\sqrt{41}}{10} t-\frac{21}{10}=-\frac{\sqrt{41}}{10}
Simplify.
t=\frac{\sqrt{41}+21}{10} t=\frac{21-\sqrt{41}}{10}
Add \frac{21}{10} to both sides of the equation.
x ^ 2 -\frac{21}{5}x +4 = 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 5
r + s = \frac{21}{5} rs = 4
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{21}{10} - u s = \frac{21}{10} + u
Two numbers r and s sum up to \frac{21}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{21}{5} = \frac{21}{10}. 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{21}{10} - u) (\frac{21}{10} + u) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
\frac{441}{100} - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-\frac{441}{100} = -\frac{41}{100}
Simplify the expression by subtracting \frac{441}{100} on both sides
u^2 = \frac{41}{100} u = \pm\sqrt{\frac{41}{100}} = \pm \frac{\sqrt{41}}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{21}{10} - \frac{\sqrt{41}}{10} = 1.460 s = \frac{21}{10} + \frac{\sqrt{41}}{10} = 2.740
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
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
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}