What are the solutions to the equation 108=-16t^2-22t+220 ?

The solutions to the equation 108=-16t^2-22t+220 are t=-(11+sqrt(1913))/(16),t=(sqrt(1913)-11)/(16)

Find the zeros of 108=-16t^2-22t+220

The zeros of 108=-16t^2-22t+220 are t=-(11+sqrt(1913))/(16),t=(sqrt(1913)-11)/(16)

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108 = - 16 t^{2} - 22 t + 220

t = - \frac{11 + 1913}{16}, t = \frac{1913 - 11}{16}

Decimal Notation

t = - 3.42111 \dots, t = 2.04611 \dots

Solution steps

108 = - 16 t^{2} - 22 t + 220

Switch sides

- 16 t^{2} - 22 t + 220 = 108

Move

108

to the left side

- 16 t^{2} - 22 t + 112 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 ( - 16 ) \cdot 112}{2 ( - 16 )}

(- 22)^{2} - 4 (- 16) \cdot 112 = 21913

t_{1, 2} = \frac{- ( - 22 ) \pm 2 1913}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 22 ) + 2 1913}{2 ( - 16 )}, t_{2} = \frac{- ( - 22 ) - 2 1913}{2 ( - 16 )}

t = \frac{- ( - 22 ) + 2 1913}{2 ( - 16 )} : - \frac{11 + 1913}{16}

t = \frac{- ( - 22 ) - 2 1913}{2 ( - 16 )} : \frac{1913 - 11}{16}

The solutions to the quadratic equation are:

t = - \frac{11 + 1913}{16}, t = \frac{1913 - 11}{16}

3 X - (2 - X) - 2 = (X + 1) (2 - 3 X)

r = r^{1} - r^{2}

4 = 1.5 x - 4.9 x^{2}

- 2 x^{2} + 50 x = 0

(n - 1) x^{2} + 5 x - n = 4 + 5 n

What are the solutions to the equation 108=-16t^2-22t+220 ?
The solutions to the equation 108=-16t^2-22t+220 are t=-(11+sqrt(1913))/(16),t=(sqrt(1913)-11)/(16)
Find the zeros of 108=-16t^2-22t+220
The zeros of 108=-16t^2-22t+220 are t=-(11+sqrt(1913))/(16),t=(sqrt(1913)-11)/(16)