What are the solutions to the equation x^2+5x+6=0.5u(y) ?

The solutions to the equation x^2+5x+6=0.5u(y) are x=(-5+sqrt(2u(y)+1))/2 ,x=-(sqrt(2u(y)+1)+5)/2

Find the zeros of x^2+5x+6=0.5u(y)

The zeros of x^2+5x+6=0.5u(y) are x=(-5+sqrt(2u(y)+1))/2 ,x=-(sqrt(2u(y)+1)+5)/2

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x^{2} + 5 x + 6 = 0.5 u (y)

x = \frac{- 5 + 2 u ( y ) + 1}{2}, x = - \frac{2 u ( y ) + 1 + 5}{2}

Solution steps

x^{2} + 5 x + 6 = 0.5 u (y)

Multiply both sides by

10

10 x^{2} + 50 x + 60 = 5 u (y)

Move

5 u (y)

to the left side

10 x^{2} + 50 x + 60 - 5 u (y) = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 \cdot 10 ( 60 - 5 u ( y ) )}{2 \cdot 10}

Simplify

5 0^{2} - 4 \cdot 10 (60 - 5 u (y)) : 10 2 u (y) + 1

x_{1, 2} = \frac{- 50 \pm 10 2 u ( y ) + 1}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 50 + 10 2 u ( y ) + 1}{2 \cdot 10}, x_{2} = \frac{- 50 - 10 2 u ( y ) + 1}{2 \cdot 10}

x = \frac{- 50 + 10 2 u ( y ) + 1}{2 \cdot 10} : \frac{- 5 + 2 u ( y ) + 1}{2}

x = \frac{- 50 - 10 2 u ( y ) + 1}{2 \cdot 10} : - \frac{2 u ( y ) + 1 + 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 2 u ( y ) + 1}{2}, x = - \frac{2 u ( y ) + 1 + 5}{2}

x = \frac{1}{9} y^{2}

a^{2} + 5 a - \frac{95}{15} = b

\frac{( y - 2 ) ^{2}}{3} = 8 - x

(y - 1)^{2} = - 8 (x + 4)

X \leq (Y + 2)^{2}

What are the solutions to the equation x^2+5x+6=0.5u(y) ?
The solutions to the equation x^2+5x+6=0.5u(y) are x=(-5+sqrt(2u(y)+1))/2 ,x=-(sqrt(2u(y)+1)+5)/2
Find the zeros of x^2+5x+6=0.5u(y)
The zeros of x^2+5x+6=0.5u(y) are x=(-5+sqrt(2u(y)+1))/2 ,x=-(sqrt(2u(y)+1)+5)/2