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

The solutions to the equation 2x^2-3x-15=4 are x=(3+sqrt(161))/4 ,x=(3-sqrt(161))/4

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

The zeros of 2x^2-3x-15=4 are x=(3+sqrt(161))/4 ,x=(3-sqrt(161))/4

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

x = \frac{3 + 161}{4}, x = \frac{3 - 161}{4}

Decimal Notation

x = 3.92214 \dots, x = - 2.42214 \dots

Solution steps

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

Move

4

to the left side

2 x^{2} - 3 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 19 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 19) = 161

x_{1, 2} = \frac{- ( - 3 ) \pm 161}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 161}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 161}{2 \cdot 2}

x = \frac{- ( - 3 ) + 161}{2 \cdot 2} : \frac{3 + 161}{4}

x = \frac{- ( - 3 ) - 161}{2 \cdot 2} : \frac{3 - 161}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 161}{4}, x = \frac{3 - 161}{4}

j^{2} + 4 j + 4 = 0

4 x^{2} = - 8 x - 2

4 x^{2} = - 8 x - 3

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

x^{2} - 3 x - 4 = - x

What are the solutions to the equation 2x^2-3x-15=4 ?
The solutions to the equation 2x^2-3x-15=4 are x=(3+sqrt(161))/4 ,x=(3-sqrt(161))/4
Find the zeros of 2x^2-3x-15=4
The zeros of 2x^2-3x-15=4 are x=(3+sqrt(161))/4 ,x=(3-sqrt(161))/4