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35x^2+16x-12=0
a = 35; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·35·(-12)
Δ = 1936
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-44}{2*35}=\frac{-60}{70} =-6/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+44}{2*35}=\frac{28}{70} =2/5 $
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