What are the solutions to the equation (4y+10)(y-3y)=5 ?

The solutions to the equation (4y+10)(y-3y)=5 are y=-(5+sqrt(15))/4 ,y=-(5-sqrt(15))/4

Find the zeros of (4y+10)(y-3y)=5

The zeros of (4y+10)(y-3y)=5 are y=-(5+sqrt(15))/4 ,y=-(5-sqrt(15))/4

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(4 y + 10) (y - 3 y) = 5

y = - \frac{5 + 15}{4}, y = - \frac{5 - 15}{4}

Solution steps

(4 y + 10) (y - 3 y) = 5

Expand

(4 y + 10) (y - 3 y) : - 8 y^{2} - 20 y

- 8 y^{2} - 20 y = 5

Move

5

to the left side

- 8 y^{2} - 20 y - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 8 ) ( - 5 )}{2 ( - 8 )}

(- 20)^{2} - 4 (- 8) (- 5) = 415

y_{1, 2} = \frac{- ( - 20 ) \pm 4 15}{2 ( - 8 )}

Separate the solutions

y_{1} = \frac{- ( - 20 ) + 4 15}{2 ( - 8 )}, y_{2} = \frac{- ( - 20 ) - 4 15}{2 ( - 8 )}

y = \frac{- ( - 20 ) + 4 15}{2 ( - 8 )} : - \frac{5 + 15}{4}

y = \frac{- ( - 20 ) - 4 15}{2 ( - 8 )} : - \frac{5 - 15}{4}

The solutions to the quadratic equation are:

y = - \frac{5 + 15}{4}, y = - \frac{5 - 15}{4}

x^{2} + 8 x - 4 = 0

3 x^{2} - 4 = - x^{2}

2 - x^{2} + x = 0

x^{2} + 9^{2} = 1 7^{2}

2 x^{2} + 4 x + 4 = 10

What are the solutions to the equation (4y+10)(y-3y)=5 ?
The solutions to the equation (4y+10)(y-3y)=5 are y=-(5+sqrt(15))/4 ,y=-(5-sqrt(15))/4
Find the zeros of (4y+10)(y-3y)=5
The zeros of (4y+10)(y-3y)=5 are y=-(5+sqrt(15))/4 ,y=-(5-sqrt(15))/4