How do I solve a quadratic equation like this: x squared minus 9x = -14?

$$x^2-9x=-14\\\\x^2-9x+14=0\\-\\-9=-7+(-2)\\14=-7\cdot(-2)\\-\\\\x^2-9x+14=(x-7)(x-2)\\\\(x-7)(x-2)=0\iff x-7=0\ or\ x-2=0\\\\x=7\ or\ x=2$$

$$x^2-9x+14=0\\\\a=1;\ b=-9;\ c=14\\\\\Delta=b^2-4ac\\\\\Delta=(-9)^2-4\cdot1\cdot14=81-56=25\\\\x_1=\frac{-b-\sqrt\Delta}{2a}\to x_1=\frac{9-\sqrt{25}}{2\cdot1}=\frac{9-5}{2}=\frac{4}{2}=2\\\\x_2=\frac{-b+\sqrt\Delta}{2a}\to x_2=\frac{9+\sqrt{25}}{2\cdot1}=\frac{9+5}{2}=\frac{14}{2}=7$$

Bring everything over to one side.
x² -9x=-14
x²-9x+14=0
Now it is an easy trinomial so you need to find two numbers that add up to -9 and multiply to equal 14. The two numbers are -7 and -2.
x²-9x+14=0
(x-7)(x-2)=0
x= 7 and 2

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