What are the solutions to the equation t^2-10t+21=0 ?

The solutions to the equation t^2-10t+21=0 are t=7,t=3

Find the zeros of t^2-10t+21=0

The zeros of t^2-10t+21=0 are t=7,t=3

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t^{2} - 10 t + 21 = 0

t = 7, t = 3

Solution steps

t^{2} - 10 t + 21 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 21}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 21 = 4

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

Separate the solutions

t_{1} = \frac{- ( - 10 ) + 4}{2 \cdot 1}, t_{2} = \frac{- ( - 10 ) - 4}{2 \cdot 1}

t = \frac{- ( - 10 ) + 4}{2 \cdot 1} : 7

t = \frac{- ( - 10 ) - 4}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

t = 7, t = 3

x^{2} + 1 = - 4 x

4 y^{2} - 12 y = 0

r^{2} + 3 r = 8

x^{2} + 8 x + 16 = 0

2 x^{2} - 50 x = 0

What are the solutions to the equation t^2-10t+21=0 ?
The solutions to the equation t^2-10t+21=0 are t=7,t=3
Find the zeros of t^2-10t+21=0
The zeros of t^2-10t+21=0 are t=7,t=3