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(6x+4)10x=95
We move all terms to the left:
(6x+4)10x-(95)=0
We multiply parentheses
60x^2+40x-95=0
a = 60; b = 40; c = -95;
Δ = b2-4ac
Δ = 402-4·60·(-95)
Δ = 24400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24400}=\sqrt{400*61}=\sqrt{400}*\sqrt{61}=20\sqrt{61}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{61}}{2*60}=\frac{-40-20\sqrt{61}}{120} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{61}}{2*60}=\frac{-40+20\sqrt{61}}{120} $
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