What are the solutions to the equation (2/3 y-24)(y-36)=240 ?

The solutions to the equation (2/3 y-24)(y-36)=240 are y=6(6+sqrt(10)),y=6(6-sqrt(10))

Find the zeros of (2/3 y-24)(y-36)=240

The zeros of (2/3 y-24)(y-36)=240 are y=6(6+sqrt(10)),y=6(6-sqrt(10))

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(\frac{2}{3} y - 24) (y - 36) = 240

y = 6 (6 + 10), y = 6 (6 - 10)

Decimal Notation

y = 54.97366 \dots, y = 17.02633 \dots

Solution steps

(\frac{2}{3} y - 24) (y - 36) = 240

Expand

(\frac{2}{3} y - 24) (y - 36) : \frac{2}{3} y^{2} - 48 y + 864

\frac{2}{3} y^{2} - 48 y + 864 = 240

Move

240

to the left side

\frac{2}{3} y^{2} - 48 y + 624 = 0

Write in the standard form

a x^{2} + b x + c = 0

\frac{2 y ^{2}}{3} - 48 y + 624 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot \frac{2}{3} \cdot 624}{2 \cdot \frac{2}{3}}

(- 48)^{2} - 4 \cdot \frac{2}{3} \cdot 624 = 810

y_{1, 2} = \frac{- ( - 48 ) \pm 8 10}{2 \cdot \frac{2}{3}}

Separate the solutions

y_{1} = \frac{- ( - 48 ) + 8 10}{2 \cdot \frac{2}{3}}, y_{2} = \frac{- ( - 48 ) - 8 10}{2 \cdot \frac{2}{3}}

y = \frac{- ( - 48 ) + 8 10}{2 \frac{2}{3}} : 6 (6 + 10)

y = \frac{- ( - 48 ) - 8 10}{2 \frac{2}{3}} : 6 (6 - 10)

The solutions to the quadratic equation are:

y = 6 (6 + 10), y = 6 (6 - 10)

81 - x^{2} = - 9

(x + 3)^{2} + (x + 2) (x - 2) = x (x - 1) - 5

u^{2} - 25 u + 144 = 0

x^{2} = 4 x + 32

x^{2} - 1 = 0

What are the solutions to the equation (2/3 y-24)(y-36)=240 ?
The solutions to the equation (2/3 y-24)(y-36)=240 are y=6(6+sqrt(10)),y=6(6-sqrt(10))
Find the zeros of (2/3 y-24)(y-36)=240
The zeros of (2/3 y-24)(y-36)=240 are y=6(6+sqrt(10)),y=6(6-sqrt(10))