What are the solutions to the equation solvefor t,s=t^2-5t+10 ?

The solutions to the equation solvefor t,s=t^2-5t+10 are t=(5+sqrt(4s-15))/2 ,t=(5-sqrt(4s-15))/2

Find the zeros of solvefor t,s=t^2-5t+10

The zeros of solvefor t,s=t^2-5t+10 are t=(5+sqrt(4s-15))/2 ,t=(5-sqrt(4s-15))/2

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solve for

t, s = t^{2} - 5 t + 10

t = \frac{5 + 4 s - 15}{2}, t = \frac{5 - 4 s - 15}{2}

Solution steps

s = t^{2} - 5 t + 10

Switch sides

t^{2} - 5 t + 10 = s

Move

s

to the left side

t^{2} - 5 t + 10 - s = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( 10 - s )}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot (10 - s) : 4 s - 15

t_{1, 2} = \frac{- ( - 5 ) \pm 4 s - 15}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 4 s - 15}{2 \cdot 1}, t_{2} = \frac{- ( - 5 ) - 4 s - 15}{2 \cdot 1}

t = \frac{- ( - 5 ) + 4 s - 15}{2 \cdot 1} : \frac{5 + 4 s - 15}{2}

t = \frac{- ( - 5 ) - 4 s - 15}{2 \cdot 1} : \frac{5 - 4 s - 15}{2}

The solutions to the quadratic equation are:

t = \frac{5 + 4 s - 15}{2}, t = \frac{5 - 4 s - 15}{2}

48 = - 16 t^{2} + 40 t

- x^{2} + 3 x + 9 = 0

x^{2} + \frac{1}{7} x - \frac{1}{7} = 0

8 n^{2} + 47 n + 50 = - 3 n

[3 (x + 10)]^{2} + [3 x]^{2} = 20 0^{2}

What are the solutions to the equation solvefor t,s=t^2-5t+10 ?
The solutions to the equation solvefor t,s=t^2-5t+10 are t=(5+sqrt(4s-15))/2 ,t=(5-sqrt(4s-15))/2
Find the zeros of solvefor t,s=t^2-5t+10
The zeros of solvefor t,s=t^2-5t+10 are t=(5+sqrt(4s-15))/2 ,t=(5-sqrt(4s-15))/2