Solve for t
t=5
t=18
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a+b=-23 ab=90
To solve the equation, factor t^{2}-23t+90 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.
-1,-90 -2,-45 -3,-30 -5,-18 -6,-15 -9,-10
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 90.
-1-90=-91 -2-45=-47 -3-30=-33 -5-18=-23 -6-15=-21 -9-10=-19
Calculate the sum for each pair.
a=-18 b=-5
The solution is the pair that gives sum -23.
\left(t-18\right)\left(t-5\right)
Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.
t=18 t=5
To find equation solutions, solve t-18=0 and t-5=0.
a+b=-23 ab=1\times 90=90
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+90. To find a and b, set up a system to be solved.
-1,-90 -2,-45 -3,-30 -5,-18 -6,-15 -9,-10
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 90.
-1-90=-91 -2-45=-47 -3-30=-33 -5-18=-23 -6-15=-21 -9-10=-19
Calculate the sum for each pair.
a=-18 b=-5
The solution is the pair that gives sum -23.
\left(t^{2}-18t\right)+\left(-5t+90\right)
Rewrite t^{2}-23t+90 as \left(t^{2}-18t\right)+\left(-5t+90\right).
t\left(t-18\right)-5\left(t-18\right)
Factor out t in the first and -5 in the second group.
\left(t-18\right)\left(t-5\right)
Factor out common term t-18 by using distributive property.
t=18 t=5
To find equation solutions, solve t-18=0 and t-5=0.
t^{2}-23t+90=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(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 90}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -23 for b, and 90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-23\right)±\sqrt{529-4\times 90}}{2}
Square -23.
t=\frac{-\left(-23\right)±\sqrt{529-360}}{2}
Multiply -4 times 90.
t=\frac{-\left(-23\right)±\sqrt{169}}{2}
Add 529 to -360.
t=\frac{-\left(-23\right)±13}{2}
Take the square root of 169.
t=\frac{23±13}{2}
The opposite of -23 is 23.
t=\frac{36}{2}
Now solve the equation t=\frac{23±13}{2} when ± is plus. Add 23 to 13.
t=18
Divide 36 by 2.
t=\frac{10}{2}
Now solve the equation t=\frac{23±13}{2} when ± is minus. Subtract 13 from 23.
t=5
Divide 10 by 2.
t=18 t=5
The equation is now solved.
t^{2}-23t+90=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}-23t+90-90=-90
Subtract 90 from both sides of the equation.
t^{2}-23t=-90
Subtracting 90 from itself leaves 0.
t^{2}-23t+\left(-\frac{23}{2}\right)^{2}=-90+\left(-\frac{23}{2}\right)^{2}
Divide -23, the coefficient of the x term, by 2 to get -\frac{23}{2}. Then add the square of -\frac{23}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-23t+\frac{529}{4}=-90+\frac{529}{4}
Square -\frac{23}{2} by squaring both the numerator and the denominator of the fraction.
t^{2}-23t+\frac{529}{4}=\frac{169}{4}
Add -90 to \frac{529}{4}.
\left(t-\frac{23}{2}\right)^{2}=\frac{169}{4}
Factor t^{2}-23t+\frac{529}{4}. 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{23}{2}\right)^{2}}=\sqrt{\frac{169}{4}}
Take the square root of both sides of the equation.
t-\frac{23}{2}=\frac{13}{2} t-\frac{23}{2}=-\frac{13}{2}
Simplify.
t=18 t=5
Add \frac{23}{2} to both sides of the equation.
x ^ 2 -23x +90 = 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 = 23 rs = 90
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{23}{2} - u s = \frac{23}{2} + u
Two numbers r and s sum up to 23 exactly when the average of the two numbers is \frac{1}{2}*23 = \frac{23}{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{23}{2} - u) (\frac{23}{2} + u) = 90
To solve for unknown quantity u, substitute these in the product equation rs = 90
\frac{529}{4} - u^2 = 90
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 90-\frac{529}{4} = -\frac{169}{4}
Simplify the expression by subtracting \frac{529}{4} on both sides
u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{23}{2} - \frac{13}{2} = 5 s = \frac{23}{2} + \frac{13}{2} = 18
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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