What are the solutions to the equation 0=t^2-9t+10 ?

The solutions to the equation 0=t^2-9t+10 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2

Find the zeros of 0=t^2-9t+10

The zeros of 0=t^2-9t+10 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2

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0 = t^{2} - 9 t + 10

t = \frac{9 + 41}{2}, t = \frac{9 - 41}{2}

Decimal Notation

t = 7.70156 \dots, t = 1.29843 \dots

Solution steps

0 = t^{2} - 9 t + 10

Switch sides

t^{2} - 9 t + 10 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 10 = 41

t_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 1}, t_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 1}

t = \frac{- ( - 9 ) + 41}{2 \cdot 1} : \frac{9 + 41}{2}

t = \frac{- ( - 9 ) - 41}{2 \cdot 1} : \frac{9 - 41}{2}

The solutions to the quadratic equation are:

t = \frac{9 + 41}{2}, t = \frac{9 - 41}{2}

(x - 2)^{2} - 4 = 2 x - 5

2 x^{2} - 20 x = - 42

xx = 6 n - 1

z^{2} + 2 z - 7 = 0

9 x^{2} - 57 x - 20 = 22

What are the solutions to the equation 0=t^2-9t+10 ?
The solutions to the equation 0=t^2-9t+10 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2
Find the zeros of 0=t^2-9t+10
The zeros of 0=t^2-9t+10 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2