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(2x+1)(x-4)-10=0
We multiply parentheses ..
(+2x^2-8x+x-4)-10=0
We get rid of parentheses
2x^2-8x+x-4-10=0
We add all the numbers together, and all the variables
2x^2-7x-14=0
a = 2; b = -7; c = -14;
Δ = b2-4ac
Δ = -72-4·2·(-14)
Δ = 161
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{161}}{2*2}=\frac{7-\sqrt{161}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{161}}{2*2}=\frac{7+\sqrt{161}}{4} $
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