# Solve the equation with the quadratic formula `x_(1,2)=(-bpm sqrt(b^2-4ac))/(2a)` and check the answer. `4b^2+5=10b`

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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.

`4b^2-10b+5=0`

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=`

`(10pm2sqrt(5))/8=(5pm sqrt5)/4`

`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

`4((5-sqrt5)/4)^2-10cdot(5-sqrt5)/4+5=4cdot(25-10sqrt5+5)/16-(5(5-sqrt5))/2+5=`

`(30-10sqrt5)/4-(25-5sqrt5)/2+5=(15-5sqrt5)/2-(25-5sqrt5)/2+5=`

`(15-5sqrt5-25+5sqrt5)/2+5=-10/2+5=-5+5=0`

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

`4((5+sqrt5)/4)^2-10cdot(5+sqrt5)/4+5=(30+10sqrt5)/4-(25+5sqrt5)/2+5=`

`(15+5sqrt5-25-5sqrt5)/2+5=-10/2+5=-5+5=0`

Which confirms that `b_2` is solution of our equation.