What are the solutions to the equation -3y^2+15y-18=0 ?

The solutions to the equation -3y^2+15y-18=0 are y=2,y=3

Find the zeros of -3y^2+15y-18=0

The zeros of -3y^2+15y-18=0 are y=2,y=3

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- 3 y^{2} + 15 y - 18 = 0

y = 2, y = 3

Solution steps

- 3 y^{2} + 15 y - 18 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 ( - 3 ) ( - 18 )}{2 ( - 3 )}

1 5^{2} - 4 (- 3) (- 18) = 3

y_{1, 2} = \frac{- 15 \pm 3}{2 ( - 3 )}

Separate the solutions

y_{1} = \frac{- 15 + 3}{2 ( - 3 )}, y_{2} = \frac{- 15 - 3}{2 ( - 3 )}

y = \frac{- 15 + 3}{2 ( - 3 )} : 2

y = \frac{- 15 - 3}{2 ( - 3 )} : 3

The solutions to the quadratic equation are:

y = 2, y = 3

8 n^{2} + 13 n = 0

2 x^{2} - 2 x - 7 = 0

\frac{6 x ^{2}}{( \frac{4 x}{y} \cdot \frac{3 y ^{2}}{2} )}

q - 2 (2 - q) \leq 0

\frac{7 π}{4} - π

What are the solutions to the equation -3y^2+15y-18=0 ?
The solutions to the equation -3y^2+15y-18=0 are y=2,y=3
Find the zeros of -3y^2+15y-18=0
The zeros of -3y^2+15y-18=0 are y=2,y=3