Let f(x)=400x3+889x2-1800.
f(1)=400+889-1800=-511; f(2)=4956,
To solve, we need to find the value c such that f(c)=0, that is, x=c is a solution.
Since f(1)<0 and f(2)>0, c must lie between 1 and 2.
We can use Newton's iterative method to find c.
xn+1=xn-f(xn)/f'(xn) which means that successive approximations of x will finally converge to a value c if we keep applying this formula. A basic calculator is all that's needed but you do need calculus to find f'(x), which is the first derivative of f(x).
f'(x)=1200x2+1778x. Now we know that c is close to 1 because there's a solution for x between 1 and 2, so we start the process with x0=1, from which we can calculate x1:
x1=x0-f(x0)/f'(x0)=1-f(1)/f'(1)=1-(-511)/(1200+1778)=3489/2978=1.17159...
Using this value for x1 we can go on to find x2, then x3 and so on until we get a value which is stable.
x2=1.15456..., x3=1.154379..., x4=1.154379203=x5. So x converges to c and, depending on how accurate we need the result to be (limited by the accuracy of the calculator). c=1.15437920336 is 11 decimal place accuracy.
But this is a cubic equation so there are two more solutions but these will be complex (contain imaginary numbers). To find these complex solutions we would use synthetic division to find the quadratic by dividing by this solution, then use the quadratic formula to solve.