What are the solutions to the equation (16(x-8)^2+25(9/4))/(400)=1 ?

The solutions to the equation (16(x-8)^2+25(9/4))/(400)=1 are x=(5sqrt(55))/8+8,x=-(5sqrt(55))/8+8

Find the zeros of (16(x-8)^2+25(9/4))/(400)=1

The zeros of (16(x-8)^2+25(9/4))/(400)=1 are x=(5sqrt(55))/8+8,x=-(5sqrt(55))/8+8

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\frac{16 ( x - 8 ) ^{2} + 25 ( \frac{9}{4} )}{400} = 1

x = \frac{5 55}{8} + 8, x = - \frac{5 55}{8} + 8

Solution steps

\frac{16 ( x - 8 ) ^{2} + 25 ( \frac{9}{4} )}{400} = 1

Multiply both sides by

400

16 (x - 8)^{2} + \frac{225}{4} = 400

Move

\frac{225}{4}

to the right side

16 (x - 8)^{2} = \frac{1375}{4}

Divide both sides by

16

(x - 8)^{2} = \frac{1375}{64}

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

x - 8 = \frac{1375}{64} : x = \frac{5 55}{8} + 8

Solve

x - 8 = - \frac{1375}{64} : x = - \frac{5 55}{8} + 8

The solutions to the quadratic equation are:

x = \frac{5 55}{8} + 8, x = - \frac{5 55}{8} + 8

4 x^{2} + 5 x + 20 = 0

12 x^{2} + 6 = 17 x

0 = - 4.9 x^{2} + 24.5 x + 9.5

y, y^{2} = 11 y - 28

1 0^{2} + x^{2} = 1 5^{2}

What are the solutions to the equation (16(x-8)^2+25(9/4))/(400)=1 ?
The solutions to the equation (16(x-8)^2+25(9/4))/(400)=1 are x=(5sqrt(55))/8+8,x=-(5sqrt(55))/8+8
Find the zeros of (16(x-8)^2+25(9/4))/(400)=1
The zeros of (16(x-8)^2+25(9/4))/(400)=1 are x=(5sqrt(55))/8+8,x=-(5sqrt(55))/8+8