To solve the equation ( 0.7x^2 = 1.3x + 2 ), we first rearrange it into standard quadratic form ( ax^2 + bx + c = 0 ).
Starting with:
[
0.7x^2 - 1.3x - 2 = 0
]
Now, we can use the quadratic formula to find the solutions for ( x ):
[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
]
Here, ( a = 0.7 ), ( b = -1.3 ), and ( c = -2 ).
Now, we can calculate ( b^2 - 4ac ):
[
b^2 = (-1.3)^2 = 1.69
]
[
4ac = 4 \times 0.7 \times -2 = -5.6
]
[
b^2 - 4ac = 1.69 - (-5.6) = 1.69 + 5.6 = 7.29
]
Now substituting into the quadratic formula:
[
x = \frac{-(-1.3) \pm \sqrt{7.29}}{2 \times 0.7}
]
[
x = \frac{1.3 \pm 2.7}{1.4}
]
This gives us two values for ( x ):
- ( x = \frac{1.3 + 2.7}{1.4} = \frac{4}{1.4} = \frac{20}{7} \approx 2.857 )
- ( x = \frac{1.3 - 2.7}{1.4} = \frac{-1.4}{1.4} = -1 )
Thus, the solutions to the equation ( 0.7x^2 = 1.3x + 2 ) are:
[
x = \frac{20}{7} \quad \text{and} \quad x = -1
]
