1) 4x^2+3x-7=0; 2) 6(2x-x)+7=0;

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1) To solve the quadratic equation 4x^2 + 3x - 7 = 0, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 4, b = 3, and c = -7. Plugging these values into the formula:

x = (-3 ± √(3^2 - 44(-7))) / 2*
x = (-3 ± √(9 + 112)) /
x = (-3 ± √121) /
x = (-3 ± 11) / 8

Therefore, the solutions are
x = (11 - 3) / 8 = 8 / 8 =
x = (-11 - 3) / 8 = -14 / 8 = -1.75

So, the solutions to the equation 4x^2 + 3x - 7 = 0 are x = 1 and x = -1.75.

2) To solve the linear equation 6(2x - x) + 7 = 0, we first simplify the expression inside the brackets:

6(2x - x) = 6(1x) = 6x

Now, the equation becomes:

6x + 7 = 0

Subtracting 7 from both sides:

6x = -7

Dividing by 6:

x = -7/6

Therefore, the solution to the equation 6(2x - x) + 7 = 0 is x = -7/6.