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When you have something like ax²+bx+c=0 You first of all need to verify if it has solutions, doing this: ∆=b²-4ac If ∆0, the equation has 2 distinct real solutions; If ∆=0, it has 2 equal real solutions (basically only one) If ∆0, it has no real solutions. Then, to solve you use the quadratic formula: x=(-b±√∆)÷2a If you have ax²+bx+c=d, just bring the d on the left and do as above, like so: ax²+bx+(c-d)=0 ∆=b²-4a(c-d) x=(-b±√∆)÷2a. I’m a bit late, hope it was usef
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