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Let's first write the equation in standard form, that is everything on the left and only 0 on the right hand side.
Now we only need to put coefficients of this equation into quadratic formula where `a=4` (coefficient by the power 2), `b=-10` (linear coefficient) and `c=5` (free coefficient). Hence we have
`b_(1,2)=(-(-10)pm sqrt((-10)^2-4cdot4cdot5))/(2cdot4)=(10pm sqrt(100-80))/(8)=(10pm sqrt(20))/8=`
`b_1=(5-sqrt5)/4` <-- First solution
`b_2=(5+sqrt5)/4` <-- Second solution
To check our solutions we simply put them into our equation instead of `b.` So for `b_1` we have
So if we put `b_1` instead of `b` into our equation we get 0 what we were supposed to get because on the right hand side we have 0. Similarly for `b_2` we have
Which confirms that `b_2` is solution of our equation.
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