What are the solutions to the equation (a-7)^2=49+4a ?

The solutions to the equation (a-7)^2=49+4a are a=18,a=0

Find the zeros of (a-7)^2=49+4a

The zeros of (a-7)^2=49+4a are a=18,a=0

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(a - 7)^{2} = 49 + 4 a

a = 18, a = 0

Solution steps

(a - 7)^{2} = 49 + 4 a

Expand

(a - 7)^{2} : a^{2} - 14 a + 49

a^{2} - 14 a + 49 = 49 + 4 a

Move

4 a

to the left side

a^{2} - 18 a + 49 = 49

Move

49

to the left side

a^{2} - 18 a = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 0 = 18

a_{1, 2} = \frac{- ( - 18 ) \pm 18}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 18 ) + 18}{2 \cdot 1}, a_{2} = \frac{- ( - 18 ) - 18}{2 \cdot 1}

a = \frac{- ( - 18 ) + 18}{2 \cdot 1} : 18

a = \frac{- ( - 18 ) - 18}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

a = 18, a = 0

(x - 3) (x - 5) = 1

(1.5 \cdot 1 0^{8}) x^{2} + x - 0.327 = 0, x

x^{2} = (32)^{2} + (32)^{2}

x^{2} - 16 + x (x - 4) = 0

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

What are the solutions to the equation (a-7)^2=49+4a ?
The solutions to the equation (a-7)^2=49+4a are a=18,a=0
Find the zeros of (a-7)^2=49+4a
The zeros of (a-7)^2=49+4a are a=18,a=0