What are the solutions to the equation 22d^2-35d+12=-3d^2 ?

The solutions to the equation 22d^2-35d+12=-3d^2 are d= 4/5 ,d= 3/5

Find the zeros of 22d^2-35d+12=-3d^2

The zeros of 22d^2-35d+12=-3d^2 are d= 4/5 ,d= 3/5

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22 d^{2} - 35 d + 12 = - 3 d^{2}

d = \frac{4}{5}, d = \frac{3}{5}

Decimal Notation

d = 0.8, d = 0.6

Solution steps

22 d^{2} - 35 d + 12 = - 3 d^{2}

Move

3 d^{2}

to the left side

25 d^{2} - 35 d + 12 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 35 ) \pm ( - 35 ) ^{2} - 4 \cdot 25 \cdot 12}{2 \cdot 25}

(- 35)^{2} - 4 \cdot 25 \cdot 12 = 5

d_{1, 2} = \frac{- ( - 35 ) \pm 5}{2 \cdot 25}

Separate the solutions

d_{1} = \frac{- ( - 35 ) + 5}{2 \cdot 25}, d_{2} = \frac{- ( - 35 ) - 5}{2 \cdot 25}

d = \frac{- ( - 35 ) + 5}{2 \cdot 25} : \frac{4}{5}

d = \frac{- ( - 35 ) - 5}{2 \cdot 25} : \frac{3}{5}

The solutions to the quadratic equation are:

d = \frac{4}{5}, d = \frac{3}{5}

(3 n - 12) x^{2} + 9 x = 18

13 - 36 x^{2} = - 12

(2 x - 3)^{2} - (x - 2) (x + 6) = (2 x + 4) (2 x - 4)

6 q^{2} - q - 15 = 0

x (3 x + 8) = 3

What are the solutions to the equation 22d^2-35d+12=-3d^2 ?
The solutions to the equation 22d^2-35d+12=-3d^2 are d= 4/5 ,d= 3/5
Find the zeros of 22d^2-35d+12=-3d^2
The zeros of 22d^2-35d+12=-3d^2 are d= 4/5 ,d= 3/5