What are the solutions to the equation (a+3)(a-5)=5 ?

The solutions to the equation (a+3)(a-5)=5 are a=1+sqrt(21),a=1-sqrt(21)

Find the zeros of (a+3)(a-5)=5

The zeros of (a+3)(a-5)=5 are a=1+sqrt(21),a=1-sqrt(21)

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(a + 3) (a - 5) = 5

a = 1 + 21, a = 1 - 21

Decimal Notation

a = 5.58257 \dots, a = - 3.58257 \dots

Solution steps

(a + 3) (a - 5) = 5

Expand

(a + 3) (a - 5) : a^{2} - 2 a - 15

a^{2} - 2 a - 15 = 5

Move

5

to the left side

a^{2} - 2 a - 20 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 20) = 221

a_{1, 2} = \frac{- ( - 2 ) \pm 2 21}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 2 ) + 2 21}{2 \cdot 1}, a_{2} = \frac{- ( - 2 ) - 2 21}{2 \cdot 1}

a = \frac{- ( - 2 ) + 2 21}{2 \cdot 1} : 1 + 21

a = \frac{- ( - 2 ) - 2 21}{2 \cdot 1} : 1 - 21

The solutions to the quadratic equation are:

a = 1 + 21, a = 1 - 21

x^{2} = 1 + x - x^{2}

- 2 x^{2} - 8 x - 5 = 0

x^{2} - 21 x + 54 = 0

a = 6 s^{2}

- 46 = 17 - c^{2}

What are the solutions to the equation (a+3)(a-5)=5 ?
The solutions to the equation (a+3)(a-5)=5 are a=1+sqrt(21),a=1-sqrt(21)
Find the zeros of (a+3)(a-5)=5
The zeros of (a+3)(a-5)=5 are a=1+sqrt(21),a=1-sqrt(21)