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

The solutions to the equation (x^2-5)/2+(2x^2-3x+7)/2 =5 are x=(3+sqrt(105))/6 ,x=(3-sqrt(105))/6

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

The zeros of (x^2-5)/2+(2x^2-3x+7)/2 =5 are x=(3+sqrt(105))/6 ,x=(3-sqrt(105))/6

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\frac{x ^{2} - 5}{2} + \frac{2 x ^{2} - 3 x + 7}{2} = 5

x = \frac{3 + 105}{6}, x = \frac{3 - 105}{6}

Decimal Notation

x = 2.20782 \dots, x = - 1.20782 \dots

Solution steps

\frac{x ^{2} - 5}{2} + \frac{2 x ^{2} - 3 x + 7}{2} = 5

Multiply both sides by

2

x^{2} - 5 + 2 x^{2} - 3 x + 7 = 10

3 x^{2} - 3 x + 2 = 10

Move

10

to the left side

3 x^{2} - 3 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

(- 3)^{2} - 4 \cdot 3 (- 8) = 105

x_{1, 2} = \frac{- ( - 3 ) \pm 105}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 105}{2 \cdot 3}, x_{2} = \frac{- ( - 3 ) - 105}{2 \cdot 3}

x = \frac{- ( - 3 ) + 105}{2 \cdot 3} : \frac{3 + 105}{6}

x = \frac{- ( - 3 ) - 105}{2 \cdot 3} : \frac{3 - 105}{6}

The solutions to the quadratic equation are:

x = \frac{3 + 105}{6}, x = \frac{3 - 105}{6}

s^{2} - s + 10 = 0

- 2 x^{2} + 28 x - 148 = 0

- 4 x^{2} + 12 x = 0

3 x^{2} - 8 x - 2 = 0

(x - 4) (x + 6) = 0

What are the solutions to the equation (x^2-5)/2+(2x^2-3x+7)/2 =5 ?
The solutions to the equation (x^2-5)/2+(2x^2-3x+7)/2 =5 are x=(3+sqrt(105))/6 ,x=(3-sqrt(105))/6
Find the zeros of (x^2-5)/2+(2x^2-3x+7)/2 =5
The zeros of (x^2-5)/2+(2x^2-3x+7)/2 =5 are x=(3+sqrt(105))/6 ,x=(3-sqrt(105))/6