What are the solutions to the equation (t-4)(t+6)=5 ?

The solutions to the equation (t-4)(t+6)=5 are t=-1+sqrt(30),t=-1-sqrt(30)

Find the zeros of (t-4)(t+6)=5

The zeros of (t-4)(t+6)=5 are t=-1+sqrt(30),t=-1-sqrt(30)

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(t - 4) (t + 6) = 5

t = - 1 + 30, t = - 1 - 30

Decimal Notation

t = 4.47722 \dots, t = - 6.47722 \dots

Solution steps

(t - 4) (t + 6) = 5

Expand

(t - 4) (t + 6) : t^{2} + 2 t - 24

t^{2} + 2 t - 24 = 5

Move

5

to the left side

t^{2} + 2 t - 29 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 29 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 29) = 230

t_{1, 2} = \frac{- 2 \pm 2 30}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 2 + 2 30}{2 \cdot 1}, t_{2} = \frac{- 2 - 2 30}{2 \cdot 1}

t = \frac{- 2 + 2 30}{2 \cdot 1} : - 1 + 30

t = \frac{- 2 - 2 30}{2 \cdot 1} : - 1 - 30

The solutions to the quadratic equation are:

t = - 1 + 30, t = - 1 - 30

x^{2} + 6 x - 22 = 0

x, x^{2} + y^{2} - 18 x + 81 = 0

X^{2} - 7 X + 12 = 0

- 9 n^{2} + 2 n + 21 = - 5 n^{2}

4 x^{2} + 6 x = 0

What are the solutions to the equation (t-4)(t+6)=5 ?
The solutions to the equation (t-4)(t+6)=5 are t=-1+sqrt(30),t=-1-sqrt(30)
Find the zeros of (t-4)(t+6)=5
The zeros of (t-4)(t+6)=5 are t=-1+sqrt(30),t=-1-sqrt(30)