What are the solutions to the equation 15x^2-23x-28=0 ?

The solutions to the equation 15x^2-23x-28=0 are x= 7/3 ,x=-4/5

Find the zeros of 15x^2-23x-28=0

The zeros of 15x^2-23x-28=0 are x= 7/3 ,x=-4/5

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15 x^{2} - 23 x - 28 = 0

x = \frac{7}{3}, x = - \frac{4}{5}

Solution steps

15 x^{2} - 23 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 23 ) \pm ( - 23 ) ^{2} - 4 \cdot 15 ( - 28 )}{2 \cdot 15}

(- 23)^{2} - 4 \cdot 15 (- 28) = 47

x_{1, 2} = \frac{- ( - 23 ) \pm 47}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- ( - 23 ) + 47}{2 \cdot 15}, x_{2} = \frac{- ( - 23 ) - 47}{2 \cdot 15}

x = \frac{- ( - 23 ) + 47}{2 \cdot 15} : \frac{7}{3}

x = \frac{- ( - 23 ) - 47}{2 \cdot 15} : - \frac{4}{5}

The solutions to the quadratic equation are:

x = \frac{7}{3}, x = - \frac{4}{5}

150 = 50 x + x^{2}

j (x) = x^{2} - x + 1, j (2 a)

x, x^{2} + 2 x^{2} y + 3 x y = 0

2 t^{2} + 3 t - 1 = 0

(x^{2} + 3 x + 2) 2 - 8 (x^{2} + 3 x) = 4

What are the solutions to the equation 15x^2-23x-28=0 ?
The solutions to the equation 15x^2-23x-28=0 are x= 7/3 ,x=-4/5
Find the zeros of 15x^2-23x-28=0
The zeros of 15x^2-23x-28=0 are x= 7/3 ,x=-4/5