What are the solutions to the equation ((3a)/8)^2+a=52 ?

The solutions to the equation ((3a)/8)^2+a=52 are a=16,a=-208/9

Find the zeros of ((3a)/8)^2+a=52

The zeros of ((3a)/8)^2+a=52 are a=16,a=-208/9

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(\frac{3 a}{8})^{2} + a = 52

a = 16, a = - \frac{208}{9}

Solution steps

(\frac{3 a}{8})^{2} + a = 52

Expand

(\frac{3 a}{8})^{2} + a : \frac{9 a ^{2}}{64} + a

\frac{9 a ^{2}}{64} + a = 52

Move

52

to the left side

\frac{9 a ^{2}}{64} + a - 52 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot \frac{9}{64} ( - 52 )}{2 \cdot \frac{9}{64}}

1^{2} - 4 \cdot \frac{9}{64} (- 52) = \frac{11}{2}

a_{1, 2} = \frac{- 1 \pm \frac{11}{2}}{2 \cdot \frac{9}{64}}

Separate the solutions

a_{1} = \frac{- 1 + \frac{11}{2}}{2 \cdot \frac{9}{64}}, a_{2} = \frac{- 1 - \frac{11}{2}}{2 \cdot \frac{9}{64}}

a = \frac{- 1 + \frac{11}{2}}{2 \frac{9}{64}} : 16

a = \frac{- 1 - \frac{11}{2}}{2 \frac{9}{64}} : - \frac{208}{9}

The solutions to the quadratic equation are:

a = 16, a = - \frac{208}{9}

4 v^{2} + 6 v + 1 = 0

(5 x + 3)^{2} = 8

6 x^{2} - 19 x - 10 = 0

10 r^{2} + 68 r - 45 = 0

(x + 1)^{2} - 2 (x - 2)^{2} = 8

What are the solutions to the equation ((3a)/8)^2+a=52 ?
The solutions to the equation ((3a)/8)^2+a=52 are a=16,a=-208/9
Find the zeros of ((3a)/8)^2+a=52
The zeros of ((3a)/8)^2+a=52 are a=16,a=-208/9