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25b^2-11=-6
We move all terms to the left:
25b^2-11-(-6)=0
We add all the numbers together, and all the variables
25b^2-5=0
a = 25; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·25·(-5)
Δ = 500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{5}}{2*25}=\frac{0-10\sqrt{5}}{50} =-\frac{10\sqrt{5}}{50} =-\frac{\sqrt{5}}{5} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{5}}{2*25}=\frac{0+10\sqrt{5}}{50} =\frac{10\sqrt{5}}{50} =\frac{\sqrt{5}}{5} $
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