What are the solutions to the equation (t+4)^2-12(t+4)+35=0 ?

The solutions to the equation (t+4)^2-12(t+4)+35=0 are t=3,t=1

Find the zeros of (t+4)^2-12(t+4)+35=0

The zeros of (t+4)^2-12(t+4)+35=0 are t=3,t=1

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(t + 4)^{2} - 12 (t + 4) + 35 = 0

t = 3, t = 1

Solution steps

(t + 4)^{2} - 12 (t + 4) + 35 = 0

Expand

(t + 4)^{2} - 12 (t + 4) + 35 : t^{2} - 4 t + 3

t^{2} - 4 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 3, t = 1

9 p^{2} - 24 p + 11 = 0

- 6 (x + 5)^{2} = 120

2 x^{2} - 30 x + 24 = 3 x - x^{2} - 48

3 x^{2} - 14 x + 12 = 0

3 x (x - 2) - x (x + 1) + 5 = 0

What are the solutions to the equation (t+4)^2-12(t+4)+35=0 ?
The solutions to the equation (t+4)^2-12(t+4)+35=0 are t=3,t=1
Find the zeros of (t+4)^2-12(t+4)+35=0
The zeros of (t+4)^2-12(t+4)+35=0 are t=3,t=1