What are the solutions to the equation 3x^2=4x-15 ?

The solutions to the equation 3x^2=4x-15 are x= 2/3+i(sqrt(41))/3 ,x= 2/3-i(sqrt(41))/3

Find the zeros of 3x^2=4x-15

The zeros of 3x^2=4x-15 are x= 2/3+i(sqrt(41))/3 ,x= 2/3-i(sqrt(41))/3

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3 x^{2} = 4 x - 15

x = \frac{2}{3} + i \frac{41}{3}, x = \frac{2}{3} - i \frac{41}{3}

Solution steps

3 x^{2} = 4 x - 15

Move

15

to the left side

3 x^{2} + 15 = 4 x

Move

4 x

to the left side

3 x^{2} + 15 - 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} - 4 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 \cdot 15}{2 \cdot 3}

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 15 : 241 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 41 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 41 i}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 41 i}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 41 i}{2 \cdot 3} : \frac{2}{3} + i \frac{41}{3}

x = \frac{- ( - 4 ) - 2 41 i}{2 \cdot 3} : \frac{2}{3} - i \frac{41}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3} + i \frac{41}{3}, x = \frac{2}{3} - i \frac{41}{3}

3 x^{2} - 243 = 0

(x + 2)^{3} (x - 1) \geq 0

u^{2} = 16

\frac{8 x h + 4 h ^{2} - 3 h}{h}

- 4 x^{4} y^{5} (- 11 x^{3} y^{3})

What are the solutions to the equation 3x^2=4x-15 ?
The solutions to the equation 3x^2=4x-15 are x= 2/3+i(sqrt(41))/3 ,x= 2/3-i(sqrt(41))/3
Find the zeros of 3x^2=4x-15
The zeros of 3x^2=4x-15 are x= 2/3+i(sqrt(41))/3 ,x= 2/3-i(sqrt(41))/3