What are the solutions to the equation (2x-5)(3x^2+7x-6)=0 ?

The solutions to the equation (2x-5)(3x^2+7x-6)=0 are x=-3,x= 5/2 ,x= 2/3

Find the zeros of (2x-5)(3x^2+7x-6)=0

The zeros of (2x-5)(3x^2+7x-6)=0 are x=-3,x= 5/2 ,x= 2/3

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(2 x - 5) (3 x^{2} + 7 x - 6) = 0

x = - 3, x = \frac{5}{2}, x = \frac{2}{3}

Decimal Notation

x = - 3, x = 2.5, x = 0.66666 \dots

Solution steps

(2 x - 5) (3 x^{2} + 7 x - 6) = 0

Factor

(2 x - 5) (3 x^{2} + 7 x - 6) : (x + 3) (2 x - 5) (3 x - 2)

(x + 3) (2 x - 5) (3 x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or 2 x - 5 = 0 or 3 x - 2 = 0

Solve

x + 3 = 0 : x = - 3

Solve

2 x - 5 = 0 : x = \frac{5}{2}

Solve

3 x - 2 = 0 : x = \frac{2}{3}

The solutions are

x = - 3, x = \frac{5}{2}, x = \frac{2}{3}

- x^{3} + 4 x^{2} - 5 x + 2 = 0

54 = a^{4}

(7^{2})^{4} = n^{8}

5 y (3 - y) (4 y + 1)^{2} = 0

x^{3} - x^{2} - x - 1 = 0

What are the solutions to the equation (2x-5)(3x^2+7x-6)=0 ?
The solutions to the equation (2x-5)(3x^2+7x-6)=0 are x=-3,x= 5/2 ,x= 2/3
Find the zeros of (2x-5)(3x^2+7x-6)=0
The zeros of (2x-5)(3x^2+7x-6)=0 are x=-3,x= 5/2 ,x= 2/3