Solve for t
t = \frac{5 \sqrt{33} - 15}{2} \approx 6.861406616
t=\frac{-5\sqrt{33}-15}{2}\approx -21.861406616
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2t^{2}+30t=300
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.
2t^{2}+30t-300=300-300
Subtract 300 from both sides of the equation.
2t^{2}+30t-300=0
Subtracting 300 from itself leaves 0.
t=\frac{-30±\sqrt{30^{2}-4\times 2\left(-300\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 30 for b, and -300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-30±\sqrt{900-4\times 2\left(-300\right)}}{2\times 2}
Square 30.
t=\frac{-30±\sqrt{900-8\left(-300\right)}}{2\times 2}
Multiply -4 times 2.
t=\frac{-30±\sqrt{900+2400}}{2\times 2}
Multiply -8 times -300.
t=\frac{-30±\sqrt{3300}}{2\times 2}
Add 900 to 2400.
t=\frac{-30±10\sqrt{33}}{2\times 2}
Take the square root of 3300.
t=\frac{-30±10\sqrt{33}}{4}
Multiply 2 times 2.
t=\frac{10\sqrt{33}-30}{4}
Now solve the equation t=\frac{-30±10\sqrt{33}}{4} when ± is plus. Add -30 to 10\sqrt{33}.
t=\frac{5\sqrt{33}-15}{2}
Divide -30+10\sqrt{33} by 4.
t=\frac{-10\sqrt{33}-30}{4}
Now solve the equation t=\frac{-30±10\sqrt{33}}{4} when ± is minus. Subtract 10\sqrt{33} from -30.
t=\frac{-5\sqrt{33}-15}{2}
Divide -30-10\sqrt{33} by 4.
t=\frac{5\sqrt{33}-15}{2} t=\frac{-5\sqrt{33}-15}{2}
The equation is now solved.
2t^{2}+30t=300
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.
\frac{2t^{2}+30t}{2}=\frac{300}{2}
Divide both sides by 2.
t^{2}+\frac{30}{2}t=\frac{300}{2}
Dividing by 2 undoes the multiplication by 2.
t^{2}+15t=\frac{300}{2}
Divide 30 by 2.
t^{2}+15t=150
Divide 300 by 2.
t^{2}+15t+\left(\frac{15}{2}\right)^{2}=150+\left(\frac{15}{2}\right)^{2}
Divide 15, the coefficient of the x term, by 2 to get \frac{15}{2}. Then add the square of \frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}+15t+\frac{225}{4}=150+\frac{225}{4}
Square \frac{15}{2} by squaring both the numerator and the denominator of the fraction.
t^{2}+15t+\frac{225}{4}=\frac{825}{4}
Add 150 to \frac{225}{4}.
\left(t+\frac{15}{2}\right)^{2}=\frac{825}{4}
Factor t^{2}+15t+\frac{225}{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{15}{2}\right)^{2}}=\sqrt{\frac{825}{4}}
Take the square root of both sides of the equation.
t+\frac{15}{2}=\frac{5\sqrt{33}}{2} t+\frac{15}{2}=-\frac{5\sqrt{33}}{2}
Simplify.
t=\frac{5\sqrt{33}-15}{2} t=\frac{-5\sqrt{33}-15}{2}
Subtract \frac{15}{2} from both sides of the equation.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
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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
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
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