What are the solutions to the equation (t+5)^2-13(t+5)+42=0 ?

The solutions to the equation (t+5)^2-13(t+5)+42=0 are t=2,t=1

Find the zeros of (t+5)^2-13(t+5)+42=0

The zeros of (t+5)^2-13(t+5)+42=0 are t=2,t=1

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(t + 5)^{2} - 13 (t + 5) + 42 = 0

t = 2, t = 1

Solution steps

(t + 5)^{2} - 13 (t + 5) + 42 = 0

Expand

(t + 5)^{2} - 13 (t + 5) + 42 : t^{2} - 3 t + 2

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

Solve with the quadratic formula

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

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

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

Separate the solutions

t_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, t_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

t = 2, t = 1

\frac{a ^{18}}{a ^{- 6}}

12 \cdot \frac{1}{8}

(- x y^{3}) (4 x^{2} y)

620 x - 2 x^{2} = 300 x + 8480

2 x^{2} - 3 x = 20

What are the solutions to the equation (t+5)^2-13(t+5)+42=0 ?
The solutions to the equation (t+5)^2-13(t+5)+42=0 are t=2,t=1
Find the zeros of (t+5)^2-13(t+5)+42=0
The zeros of (t+5)^2-13(t+5)+42=0 are t=2,t=1